Differential Equations: Chapter 9 Links
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Differential Equations Exercise 9.4 introduces Homogeneous Differential Equations and the substitution method used to solve them. In this exercise, you will learn to identify homogeneous differential equations and transform them into Variable Separable form using a suitable substitution.
\displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right)\;\Longrightarrow\;y=vx\;\Longrightarrow\;\frac{dy}{dx}=v+x\frac{dv}{dx}
The key idea is to recognise equations in which the right-hand side can be expressed as a function of y/x. By making the substitution y = vx (or x = vy when the equation is given in terms of dx/dy), the differential equation is converted into a Variable Separable equation, which can then be solved by separating the variables and integrating both sides.
By practising these questions, you will learn to identify homogeneous differential equations, choose the correct substitution, and solve them confidently. These concepts are important for the CBSE board examinations and CUET.
Key Concepts
Before solving Exercise 9.4, it is important to understand what a Homogeneous Differential Equation is and how it can be transformed into a Variable Separable equation using a suitable substitution.
1. Homogeneous Function
A function F(x, y) is called a homogeneous function of degree n if replacing x by λx and y by λy gives
\displaystyle F(\lambda x,\lambda y)=\lambda^nF(x,y)
where λ is any non-zero constant and n is called the degree of the homogeneous function.
2. Homogeneous Differential Equation
A differential equation is said to be homogeneous if it can be written in either of the following forms:
\displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right)\qquad\text{or}\qquad \frac{dx}{dy}=f\!\left(\frac{x}{y}\right)
That is, the right-hand side depends only on the ratio y/x (or x/y when the equation is given in terms of dx/dy).
3. How to Identify a Homogeneous Differential Equation
- Method 1: Check whether the function satisfies \displaystyle F(\lambda x,\lambda y)=F(x,y). If yes, it is homogeneous of degree 0.
- Method 2: Rewrite the right-hand side as a function of \displaystyle \frac{y}{x} (or \displaystyle \frac{x}{y}). If this is possible, the differential equation is homogeneous.
4. Solving a Homogeneous Differential Equation
For an equation of the form
\displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right)
make the substitution
\displaystyle y=vx
Then
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting this into the differential equation converts it into a Variable Separable equation, which can then be solved by separating the variables and integrating both sides.
5. When to Use x = vy
If the differential equation is given in the form
\displaystyle \frac{dx}{dy}=f\!\left(\frac{x}{y}\right)
use the substitution \displaystyle x=vy. The equation is then solved in the same way by separating the variables and integrating both sides.
6. Important Points
- First check whether the differential equation is homogeneous before choosing a substitution.
- Use y = vx for equations involving dy/dx and x = vy for equations involving dx/dy.
- After substitution, separate the variables completely before integrating.
- Always include the constant of integration C in the final answer.
Before You Begin… Before applying the substitution, first check whether the differential equation is homogeneous. If the right-hand side can be written as a function of y/x, use the substitution y = vx. If the equation is in terms of dx/dy, use x = vy instead.
Let us now solve all the NCERT questions step by step of Differential Equations 9.4.
Question 1: Homogeneous Differential Equations
1. Show that the following differential equation is homogeneous and solve it:
\displaystyle (x^2+xy)\,dy=(x^2+y^2)\,dx
Solution
Think First… Rewrite the equation in the form \displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right). Since the right-hand side depends only on y/x, the equation is homogeneous. Then use the substitution y=vx.
We have, \displaystyle (x^2+xy)\,dy=(x^2+y^2)\,dx
Dividing both sides by \displaystyle dx(x^2+xy), we get
\displaystyle \frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}
Dividing the numerator and denominator by \displaystyle x^2,
\displaystyle \frac{dy}{dx}=\frac{1+\left(\frac{y}{x}\right)^2}{1+\frac{y}{x}}\qquad...(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting the values of y and \displaystyle \frac{dy}{dx} into the homogeneous differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}
Subtracting \displaystyle v from both sides,
We get, \displaystyle x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v
\displaystyle =\frac{1+v^2-v(1+v)}{1+v}
\displaystyle =\frac{1-v}{1+v}
Separating the variables, we get
\displaystyle \frac{1+v}{1-v}\,dv=\frac{dx}{x}\qquad...(2)
Rewriting the numerator on LHS as \displaystyle -(1-v)+2 and splitting,
We get, \displaystyle \frac{1+v}{1-v}=\frac{-(1-v)+2}{1-v}
\displaystyle =-\frac{1-v}{1-v}+\frac{2}{1-v}
\displaystyle =-1+\frac{2}{1-v}
Substituting this value on LHS of equation (2) and integrating both sides, we get
\displaystyle \int\left(-1+\frac{2}{1-v}\right)dv=\int\frac{dx}{x}
\displaystyle \Rightarrow -v-2\log|1-v|=\log|x|+C
Replacing \displaystyle v=\frac{y}{x}, we get
\displaystyle -\frac{y}{x}-2\log\left|1-\frac{y}{x}\right|=\log|x|+C
or \displaystyle -\frac{y}{x}-2\log\left|\frac{x-y}{x}\right|=\log|x|+C
Using the property \displaystyle \log\frac{a}{b}=\log a-\log b, we get
\displaystyle -\frac{y}{x}-2\left(\log|x-y|-\log|x|\right)=\log|x|+C
or \displaystyle -\frac{y}{x}-2\log|x-y|+2\log|x|=\log|x|+C
Simplifying, we get
\displaystyle 2\log|x-y|=\log|x|-\frac{y}{x}+C
Writing \displaystyle \frac{y}{x} and constant in terms of logarithms to combine,
\displaystyle 2\log|x-y|=\log|x|+\log\!\left(e^{-y/x}\right)+\log C
Using log properties on both sides,
\displaystyle \log(x-y)^2=\log\!\left(Cxe^{-y/x}\right)
Taking anti-logs on both sides, we get
\displaystyle (x-y)^2=Cxe^{-y/x}
Hence, the required general solution is
\boxed{\displaystyle (x-y)^2=Cxe^{-y/x}}
Question 2: Differential Equations 9.4
2. Show that the following differential equation is homogeneous and solve it:
\displaystyle \frac{dy}{dx}=\frac{x+y}{x}
Solution
Think First… Rewrite the right-hand side as a function of y/x. Since it depends only on y/x, the equation is homogeneous. Then use the substitution y=vx.
We have, \displaystyle \frac{dy}{dx}=\frac{x+y}{x}
Writing the right-hand side in terms of \displaystyle \frac{y}{x}, we get
\displaystyle \frac{dy}{dx}=1+\frac{y}{x} …(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Alternatively, define
\displaystyle F(x,y)=\frac{x+y}{x}
Now,
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}=F(x,y)
Hence, F(x,y) is homogeneous of degree 0. Therefore, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting \displaystyle y=vx and \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} into equation (1), we get
\displaystyle v+x\frac{dv}{dx}=1+v
Subtracting \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=1
Separating the variables, we get
\displaystyle dv=\frac{dx}{x}
Integrating both sides, we get
\displaystyle v=\log|x|+C
Replacing \displaystyle v=\frac{y}{x}, we get
\displaystyle \frac{y}{x}=\log|x|+C
Multiplying both sides by \displaystyle x, we obtain
\displaystyle y=x\log|x|+Cx
Hence, the required general solution is
\boxed{\displaystyle y=x\log|x|+Cx}
You may already be following Maths Better for the NCERT Solutions of the previous Class 12 Maths chapters, namely:
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
Likewise, Differential Equations 9.4 helps you master the method of solving Homogeneous Differential Equations. With sufficient practice, you will learn to identify homogeneous equations, choose the correct substitution and reduce them to variable separable form with confidence. This technique forms the foundation for solving many important first-order differential equations. Now, let’s proceed to the next question.
Question 3: Differential Equations
3. Show that the following differential equation is homogeneous and solve it:
\displaystyle (x-y)\,dy-(x+y)\,dx=0
Solution
Think First… Rewrite the equation in the form \displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right). Since the right-hand side depends only on y/x, the equation is homogeneous. Then use the substitution y=vx.
We have, \displaystyle (x-y)\,dy=(x+y)\,dx
Dividing both sides by \displaystyle dx(x-y), we get
\displaystyle \frac{dy}{dx}=\frac{x+y}{x-y}
Dividing the numerator and denominator by \displaystyle x on RHS, we get
\displaystyle \frac{dy}{dx}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}\qquad...(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Alternatively, define
\displaystyle F(x,y)=\frac{x+y}{x-y}
Now,
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}=F(x,y)
Hence, F(x,y) is homogeneous of degree 0. Therefore, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting \displaystyle y=vx and \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} into equation (1), we get
\displaystyle v+x\frac{dv}{dx}=\frac{1+v}{1-v}
Subtracting \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=\frac{1+v}{1-v}-v
Taking LCM on the RHS, we get
\displaystyle x\frac{dv}{dx}=\frac{1+v-v(1-v)}{1-v}
\displaystyle x\frac{dv}{dx}=\frac{1+v-v+v^2}{1-v}
This simplifies to
\displaystyle x\frac{dv}{dx}=\frac{1+v^2}{1-v}
Separating the variables, we get
\displaystyle \frac{1-v}{1+v^2}\,dv=\frac{dx}{x}
Splitting the numerator on LHS, we get
\displaystyle \left(\frac{1}{1+v^2}-\frac{v}{1+v^2}\right)dv=\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\frac{dv}{1+v^2}-\int\frac{v}{1+v^2}\,dv=\int\frac{dx}{x}\qquad...(2)
Using the standard integral formulas
\displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x+C\qquad\text{and}\qquad\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C
where \displaystyle f(v)=1+v^2 and \displaystyle f'(v)=2v, we get from equation (2)
\displaystyle \tan^{-1}v-\frac12\log(1+v^2)=\log|x|+C
Replacing \displaystyle v=\frac{y}{x}, we get
\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)-\frac12\log\!\left(1+\frac{y^2}{x^2}\right)=\log|x|+C
or \displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)-\frac12\log\!\left(\frac{x^2+y^2}{x^2}\right)=\log|x|+C
Using the property \displaystyle \log\frac{a}{b}=\log a-\log b, we get
\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)-\frac12\left(\log(x^2+y^2)-\log x^2\right)=\log|x|+C
Expanding the brackets, we get
\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)-\frac12\log(x^2+y^2)+\frac12\log x^2=\log|x|+C
Since \displaystyle \log x^2=2\log|x|, we get
\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)-\frac12\log(x^2+y^2)+\log|x|=\log|x|+C
Cancelling \displaystyle \log|x| from both sides, we obtain
\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)-\frac12\log(x^2+y^2)=C
or \displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)=\frac12\log(x^2+y^2)+C
Hence, the required general solution is
\boxed{\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)=\frac12\log(x^2+y^2)+C}
Question 4: Differential Equations 9.4
4. Show that the following differential equation is homogeneous and solve it:
\displaystyle (x^2-y^2)\,dx+2xy\,dy=0
Solution
Think First… Rewrite the equation in the form \displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right). Since the right-hand side depends only on y/x, the equation is homogeneous. Then use the substitution y=vx.
We have, \displaystyle 2xy\,dy=-(x^2-y^2)\,dx
Dividing both sides by \displaystyle 2xydx, we get
\displaystyle \frac{dy}{dx}=\frac{y^2-x^2}{2xy}
Dividing the numerator and denominator by \displaystyle x^2,
\displaystyle \frac{dy}{dx}=\frac{\left(\frac{y}{x}\right)^2-1}{2\left(\frac{y}{x}\right)}\qquad...(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Alternatively, define
\displaystyle F(x,y)=\frac{y^2-x^2}{2xy}
Now,
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda^2y^2-\lambda^2x^2}{2\lambda x\lambda y}=\frac{y^2-x^2}{2xy}=F(x,y)
Hence, F(x,y) is homogeneous of degree 0. Therefore, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting \displaystyle y=vx and \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} into equation (1), we get
\displaystyle v+x\frac{dv}{dx}=\frac{v^2-1}{2v}
Subtracting \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=\frac{v^2-1}{2v}-v
Taking LCM on the RHS,
\displaystyle x\frac{dv}{dx}=\frac{v^2-1-2v^2}{2v}
Simplifying the numerator, we get
\displaystyle x\frac{dv}{dx}=-\frac{v^2+1}{2v}
Separating the variables, we get
\displaystyle \frac{2v}{v^2+1}\,dv=-\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\frac{2v}{v^2+1}\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow \log(v^2+1)=-\log|x|+C
Replacing \displaystyle v=\frac{y}{x}, we get
\displaystyle \log\!\left(1+\frac{y^2}{x^2}\right)=-\log|x|+C
i.e. \displaystyle \log\!\left(\frac{x^2+y^2}{x^2}\right)=-\log|x|+C
Using the property \displaystyle \log\frac{a}{b}=\log a-\log b, we get
\displaystyle \log(x^2+y^2)-\log x^2=-\log|x|+C
Since \displaystyle \log x^2=2\log|x|, we get
\displaystyle \log(x^2+y^2)-2\log|x|=-\log|x|+C
\displaystyle \Rightarrow \log(x^2+y^2)=\log|x|+C
Writing the constant in logarithmic form and combining the logarithms, we get
\displaystyle \log(x^2+y^2)=\log(C|x|)
Taking anti-logs on both sides, we get
\displaystyle x^2+y^2=Cx
Hence, the required general solution is
\boxed{\displaystyle x^2+y^2=Cx}
Maths Better Tip… While practising NCERT questions, always try to simplify your final answer so that it matches the textbook answer. Whenever logarithmic terms and the arbitrary constant appear together, you may write the constant as \displaystyle \log C (where \displaystyle C>0) and then combine all the logarithmic terms using the properties of logarithms. Likewise, whenever exponential or logarithmic expressions can be simplified into an equivalent form, do so. These small simplifications often help your final answer match the NCERT answer exactly.
Question 5: Differential Equations 9.4
5. Show that the following differential equation is homogeneous and solve it:
\displaystyle x^2\frac{dy}{dx}=x^2-2y^2+xy
Solution
Think First… Rewrite the equation in the form \displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right). Since the right-hand side depends only on y/x, the equation is homogeneous. Then use the substitution y=vx.
We have, \displaystyle x^2\frac{dy}{dx}=x^2-2y^2+xy
Dividing both sides by \displaystyle x^2, we get
\displaystyle \frac{dy}{dx}=1-2\left(\frac{y}{x}\right)^2+\frac{y}{x}\qquad...(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Alternatively, define
\displaystyle F(x,y)=1-2\left(\frac{y}{x}\right)^2+\frac{y}{x}
Now,
\displaystyle F(\lambda x,\lambda y)=1-2\left(\frac{\lambda y}{\lambda x}\right)^2+\frac{\lambda y}{\lambda x}=F(x,y)
Hence, F(x,y) is homogeneous of degree 0. Therefore, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting \displaystyle y=vx and \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} into equation (1), we get
\displaystyle v+x\frac{dv}{dx}=1-2v^2+v
Subtracting \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=1-2v^2
Taking \displaystyle -2 common from the right-hand side, we get
\displaystyle x\frac{dv}{dx}=-2\left(v^2-\frac12\right)
Separating the variables, we get
\displaystyle \frac{dv}{v^2-\frac12}=-2\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\frac{dv}{v^2-\frac12}=-2\int\frac{dx}{x}\qquad...(2)
Using the special integral formula
\displaystyle \int\frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x-a}{x+a}\right|+C
with \displaystyle a=\frac1{\sqrt2}, equation (2) gives
\displaystyle \frac1{\sqrt2}\log\left|\frac{v-\frac1{\sqrt2}}{v+\frac1{\sqrt2}}\right|=-2\log|x|+C
Replacing \displaystyle v=\frac{y}{x}, we get
\displaystyle \frac1{\sqrt2}\log\left|\frac{\frac{y}{x}-\frac1{\sqrt2}}{\frac{y}{x}+\frac1{\sqrt2}}\right|=-2\log|x|+C
Taking LCM inside the log term on the LHS, we get
\displaystyle \frac1{\sqrt2}\log\left|\frac{\sqrt2\,y-x}{\sqrt2\,y+x}\right|=-2\log|x|+C
(This form is mathematically correct, and you may leave your answer here. However, to match the NCERT answer exactly, let us simplify it further.)
Multiplying the numerator and denominator inside the logarithm by \displaystyle -1, we get
\displaystyle \frac1{\sqrt2}\log\left|\frac{x-\sqrt2\,y}{x+\sqrt2\,y}\right|=-2\log|x|+C
Multiplying both sides by \displaystyle -\frac12 and using log properties, we obtain
\displaystyle \frac1{2\sqrt2}\log\left|\frac{x+\sqrt2\,y}{x-\sqrt2\,y}\right|=\log|x|+C
Hence, the required general solution is
\boxed{\displaystyle \frac1{2\sqrt2}\log\left|\frac{x+\sqrt2\,y}{x-\sqrt2\,y}\right|=\log|x|+C}
Maths Better Tip! While simplifying a solution, you may notice that we multiply or divide both sides by different non-zero constants but continue to write the arbitrary constant simply as \displaystyle C. This is perfectly correct because any non-zero multiple of an arbitrary constant is itself an arbitrary constant. If you prefer, you may rename it as \displaystyle C_1,\;C_2,\;\text{etc.} after each such step. Both approaches are mathematically correct. In fact, both methods are used in the next question below.
Question 6: Exercise 9.4
6. Show that the following differential equation is homogeneous and solve it:
\displaystyle x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx
Solution
Think First… Rewrite the equation in the form \displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right). Since the right-hand side depends only on y/x, the equation is homogeneous. Then use the substitution y=vx.
We have, \displaystyle x\,dy=y\,dx+\sqrt{x^2+y^2}\,dx
Dividing both sides by \displaystyle x\,dx, we get
\displaystyle \frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x}
Writing the square root in terms of \displaystyle \frac{y}{x}, we get
\displaystyle \frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}\qquad...(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Alternatively, define
\displaystyle F(x,y)=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}
Now,
\displaystyle F(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}+\sqrt{1+\left(\frac{\lambda y}{\lambda x}\right)^2}=F(x,y)
Hence, F(x,y) is homogeneous of degree 0. Therefore, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting \displaystyle y=vx and \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} into equation (1), we get
\displaystyle v+x\frac{dv}{dx}=v+\sqrt{1+v^2}
Subtracting \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=\sqrt{1+v^2}
Separating the variables, we get
\displaystyle \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\frac{dv}{\sqrt{1+v^2}}=\int\frac{dx}{x}\qquad...(2)
Using the special integral formula
\displaystyle \int\frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C
with \displaystyle a=1, equation (2) becomes
\displaystyle \log\left|v+\sqrt{1+v^2}\right|=\log|x|+C
Replacing \displaystyle v=\frac{y}{x}, we get
\displaystyle \log\left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log|x|+C
Since
\displaystyle \sqrt{1+\frac{y^2}{x^2}}=\sqrt{\frac{x^2+y^2}{x^2}}=\frac{\sqrt{x^2+y^2}}{|x|}
Taking the LCM inside the log term on LHS, we get
\displaystyle \log\left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=\log|x|+C
Using the property \displaystyle \log\frac{a}{b}=\log a-\log b, we get
\displaystyle \log\left|y+\sqrt{x^2+y^2}\right|-\log|x|=\log|x|+C
or \displaystyle \log\left|y+\sqrt{x^2+y^2}\right|=2\log|x|+C
Writing \displaystyle C=\log C_1, where \displaystyle C_1>0, we get
\displaystyle \log\left|y+\sqrt{x^2+y^2}\right|=\log x^2+\log C_1
Combining the logarithms on the right-hand side, we get
\displaystyle \log\left|y+\sqrt{x^2+y^2}\right|=\log(C_1x^2)
Taking antilogarithm on both sides, we get
\displaystyle y+\sqrt{x^2+y^2}=\pm C_1x^2
Since the arbitrary constant can be positive or negative, we may replace \displaystyle \pm C_1 by another arbitrary constant \displaystyle C.
Hence, the required general solution is
\boxed{\displaystyle y+\sqrt{x^2+y^2}=Cx^2}
Maths Better Tip… While simplifying the final answer, you may replace the arbitrary constant by another arbitrary constant whenever convenient. For example, if an equation contains logarithms, you may write the constant as \displaystyle \log C. To be very precise, you could write it as \displaystyle \log C_1, since it represents a new arbitrary constant. However, both notations are mathematically correct because \displaystyle C and \displaystyle C_1 are simply arbitrary constants. Similarly, constants may also absorb factors or signs during simplification, provided they remain arbitrary.
Question 7: Differential Equations 9.4
7. Show that the following differential equation is homogeneous and solve it:
\displaystyle \left\{x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right\}y\,dx=\left\{y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right\}x\,dy
Solution
Think First… Rewrite the equation in the form \displaystyle \frac{dy}{dx}=f\!\left(\frac{y}{x}\right). Then use the substitution y=vx.
We have,
\displaystyle \left\{x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right\}y\,dx=\left\{y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right\}x\,dy
Rewriting to collect \displaystyle \frac{dy}{dx} on one side, we get
\displaystyle \frac{dy}{dx}=\frac{y\left\{x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right\}}{x\left\{y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right\}}
Dividing the numerator and denominator on the right-hand side by \displaystyle x, we get
\displaystyle \frac{dy}{dx}=\frac{\frac{y}{x}\left\{\cos\left(\frac{y}{x}\right)+\frac{y}{x}\sin\left(\frac{y}{x}\right)\right\}}{\frac{y}{x}\sin\left(\frac{y}{x}\right)-\cos\left(\frac{y}{x}\right)} …(1)
Since the right-hand side is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Let \displaystyle y=vx
Then, \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting \displaystyle y=vx and \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} into equation (1), we get
\displaystyle v+x\frac{dv}{dx}=\frac{v(\cos v+v\sin v)}{v\sin v-\cos v}
Taking \displaystyle v to the right-hand side, we get
\displaystyle x\frac{dv}{dx}=\frac{v(\cos v+v\sin v)}{v\sin v-\cos v}-v
Taking LCM on the right-hand side, we get
\displaystyle x\frac{dv}{dx}=\frac{v(\cos v+v\sin v)-v(v\sin v-\cos v)}{v\sin v-\cos v}
Simplifying the numerator, we get
\displaystyle x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}
Separating the variables, we get
\displaystyle \frac{v\sin v-\cos v}{2v\cos v}\,dv=\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\frac{v\sin v-\cos v}{2v\cos v}\,dv=\int\frac{dx}{x}
Splitting the integral on the left-hand side, we obtain
\displaystyle \frac12\int\tan v\,dv-\frac12\int\frac{dv}{v}=\int\frac{dx}{x}
\displaystyle -\frac12\log|\cos v|-\frac12\log|v|=\log|x|+\log C
Combining the log terms on the left-hand side using log properties, we get
\displaystyle -\frac12\log|v\cos v|=\log|x|+\log C
or \displaystyle \log|v\cos v|=-2\log|x|+\log C
Substituting \displaystyle v=\frac{y}{x}, we get
\displaystyle \log\left|\frac{y}{x}\cos\left(\frac{y}{x}\right)\right|=-2\log|x|+\log C
Adding \displaystyle \log|x| to both sides and using log property, we get
\displaystyle \log\left|y\cos\left(\frac{y}{x}\right)\right|=-\log|x|+\log C
Taking the antilogarithm of both sides, we get
\displaystyle y\cos\left(\frac{y}{x}\right)=\frac{C}{x}
or \displaystyle xy\cos\left(\frac{y}{x}\right)=C
Hence, the required general solution is
\boxed{\displaystyle xy\cos\left(\frac{y}{x}\right)=C}
By now, you have already completed the NCERT Solutions for Matrices, Determinants, Relations and Functions, Inverse Trigonometric Functions, Continuity and Differentiability, Application of Derivatives, Integrals and Application of Integrals. Likewise, every exercise of Differential Equations is explained here with detailed, step-by-step solutions to help you build strong concepts and prepare confidently for your CBSE board examinations.
Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, where each solution is explained in a simple and student-friendly manner. Now, let’s proceed to the next question.
Question 8: Differential Equations 9.4
8. Show that the following differential equation is homogeneous and solve it:
\displaystyle x\frac{dy}{dx}-y+x\sin\left(\frac{y}{x}\right)=0
Solution
Think First… Write the equation in the form \displaystyle \frac{dy}{dx}. Since the RHS depends only on \displaystyle \frac{y}{x}, the equation is homogeneous. Then substitute \displaystyle y=vx and separate the variables.
We have, \displaystyle x\frac{dy}{dx}-y+x\sin\left(\frac{y}{x}\right)=0
Rewriting the equation to collect \displaystyle \frac{dy}{dx} on one side and dividing by x, we get
\displaystyle \frac{dy}{dx}=\frac{y}{x}-\sin\left(\frac{y}{x}\right)\qquad...(1)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=v-\sin v
or \displaystyle x\frac{dv}{dx}=-\sin v
Separating the variables, we get
\displaystyle \frac{dv}{\sin v}=-\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\cosec v\,dv=-\int\frac{dx}{x}\qquad...(2)
Using the standard integral formula
\displaystyle \int\cosec x\,dx=\log|\cosec x-\cot x|+C
we get from equation (2)
\displaystyle \log|\cosec v-\cot v|=-\log|x|+\log C
Substituting \displaystyle v=\frac{y}{x}, we get
\displaystyle \log\left|\cosec\left(\frac{y}{x}\right)-\cot\left(\frac{y}{x}\right)\right|=-\log|x|+\log C
Combining the logarithms on the right-hand side, we get
\displaystyle \log\left|\cosec\left(\frac{y}{x}\right)-\cot\left(\frac{y}{x}\right)\right|=\log\left(\frac{C}{x}\right)
Taking antilogarithm on both sides, we get
\displaystyle \cosec\left(\frac{y}{x}\right)-\cot\left(\frac{y}{x}\right)=\frac{C}{x}
Using the trigonometric identity
\displaystyle \cosec\theta-\cot\theta=\frac{1-\cos\theta}{\sin\theta}
with \displaystyle \theta=\frac{y}{x}, we get
\displaystyle \frac{1-\cos\left(\frac{y}{x}\right)}{\sin\left(\frac{y}{x}\right)}=\frac{C}{x}
Cross-multiplying, we get
\displaystyle x\left[1-\cos\left(\frac{y}{x}\right)\right]=C\sin\left(\frac{y}{x}\right)
Hence, the required general solution is
\boxed{\displaystyle x\left[1-\cos\left(\frac{y}{x}\right)\right]=C\sin\left(\frac{y}{x}\right)}
Question 9: Differential Equations 9.4
9. Show that the following differential equation is homogeneous and solve it:
\displaystyle y\,dx+x\log\left(\frac{y}{x}\right)dy-2x\,dy=0
Solution
Think First… Write the equation in the form \displaystyle \frac{dy}{dx}. Since the RHS depends only on \displaystyle \frac{y}{x}, substitute \displaystyle y=vx and simplify carefully.
We have,
\displaystyle y\,dx+x\log\left(\frac{y}{x}\right)dy-2x\,dy=0
Rewriting the equation to collect \displaystyle \frac{dy}{dx} on one side and dividing by x, we get
\displaystyle \frac{dy}{dx}=\frac{\frac{y}{x}}{2-\log\left(\frac{y}{x}\right)}\qquad...(1)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=\frac{v}{2-\log v}
or \displaystyle x\frac{dv}{dx}=\frac{v}{2-\log v}-v
Taking LCM on the right-hand side, we get
\displaystyle x\frac{dv}{dx}=\frac{v-v(2-\log v)}{2-\log v}
Simplifying the numerator, we get
\displaystyle x\frac{dv}{dx}=\frac{v(\log v-1)}{2-\log v}
Separating the variables, we get
\displaystyle \frac{2-\log v}{v(\log v-1)}\,dv=\frac{dx}{x}
💡(Maths Better Note: Sometimes, an integral cannot be evaluated directly. In such cases, first modify or rewrite the integrand into a form whose standard integral is known. This is the same idea we used in Exercise 7.5 while rewriting rational functions before integration. Recognizing such patterns becomes easier with practice.)
Writing \displaystyle 2-\log v=-(\log v-1)+1, we get
\displaystyle \int\frac{-(\log v-1)+1}{v(\log v-1)}\,dv=\int\frac{dx}{x}
Splitting the integral on the left-hand side, we get
\displaystyle -\int\frac{dv}{v}+\int\frac{dv}{v(\log v-1)}=\int\frac{dx}{x}\qquad...(2)
For the second integral, let
\displaystyle t=\log v-1 \Rightarrow dt=\frac{dv}{v}
Thus, equation (2) becomes
\displaystyle -\log|v|+\int\frac{dt}{t}=\log|x|+\log C
or \displaystyle -\log|v|+\log|t|=\log|x|+\log C
Substituting \displaystyle t=\log v-1 and using log properties, we get
\displaystyle \log\left|\frac{\log v-1}{v}\right|=\log|Cx|
Taking antilogarithm on both sides, we get
\displaystyle \frac{\log v-1}{v}=Cx
Substituting \displaystyle v=\frac{y}{x}, we get
\displaystyle \frac{\log\left(\frac{y}{x}\right)-1}{\frac{y}{x}}=Cx
Cross-multiplying, we get
\displaystyle \log\left(\frac{y}{x}\right)-1=Cy
Hence, the required general solution is
\boxed{\displaystyle Cy=\log\left(\frac{y}{x}\right)-1}
Question 10: Differential Equations Exercise 9.4
10. Show that the following differential equation is homogeneous and solve it:
\displaystyle \left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0
Solution
Think First… Since \displaystyle e^{\frac{x}{y}} contains the ratio \displaystyle \frac{x}{y}, it is convenient to write the equation in the form \displaystyle \frac{dx}{dy}. Then substitute \displaystyle x=vy and separate the variables.
We have,
\displaystyle \left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0
Rewriting the equation to collect \displaystyle \frac{dx}{dy}, we get
\displaystyle \frac{dx}{dy}=-\frac{e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}\qquad...(1)
Since the RHS is a function of \displaystyle \frac{x}{y} only, the given differential equation is homogeneous.
Substituting \displaystyle x=vy, we have
\displaystyle \frac{dx}{dy}=v+y\frac{dv}{dy}
Substituting these values into the differential equation (1), we get
\displaystyle v+y\frac{dv}{dy}=-\frac{e^v(1-v)}{1+e^v}
or \displaystyle y\frac{dv}{dy}=-\frac{e^v(1-v)}{1+e^v}-v
Taking LCM on the right-hand side, we get
\displaystyle y\frac{dv}{dy}=\frac{-e^v(1-v)-v(1+e^v)}{1+e^v}
Simplifying the numerator, we get
\displaystyle y\frac{dv}{dy}=-\frac{e^v+v}{1+e^v}
Separating the variables, we get
\displaystyle \frac{1+e^v}{e^v+v}\,dv=-\frac{dy}{y}
Integrating both sides, we get
\displaystyle \int\frac{1+e^v}{e^v+v}\,dv=-\int\frac{dy}{y}\qquad...(2)
Since \displaystyle \frac{d}{dv}(e^v+v)=1+e^v, using the standard integral formula
\displaystyle \int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C
we get from equation (2)
\displaystyle \log|e^v+v|=-\log|y|+\log C
Combining the logarithms, we get
\displaystyle \log|y(e^v+v)|=\log C
Taking antilogarithm on both sides, we get
\displaystyle y(e^v+v)=C
Substituting \displaystyle v=\frac{x}{y}, we get
\displaystyle y\left(e^{\frac{x}{y}}+\frac{x}{y}\right)=C
Simplifying, we get
\boxed{\displaystyle x+ye^{\frac{x}{y}}=C}
Maths Better Tip! Before solving a homogeneous differential equation, first decide whether it is easier to work with \displaystyle \frac{dy}{dx} or \displaystyle \frac{dx}{dy}. Then choose the substitution accordingly:
- If you work with \displaystyle \frac{dy}{dx}, use \displaystyle y=vx.
- If you work with \displaystyle \frac{dx}{dy}, use \displaystyle x=vy.
Sometimes the better choice is obvious, while sometimes it comes with practice or a little trial and error. However, if the equation already contains the ratio \displaystyle \frac{y}{x} or \displaystyle \frac{x}{y} inside logarithms, trigonometric functions or exponents, it usually suggests the substitution to use.
Question 11: Particular Solution
11. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle (x+y)\,dy+(x-y)\,dx=0;\; y=1\text{ when }x=1
Solution
Think First… Since this is a particular solution, first find the general solution of the differential equation. Then use the given condition to determine the arbitrary constant. Write the equation in the form \displaystyle \frac{dy}{dx}. Since the RHS is a function of \displaystyle \frac{y}{x} only, substitute \displaystyle y=vx.
We have,
\displaystyle (x+y)\,dy+(x-y)\,dx=0
Rewrite the equation to collect \displaystyle \frac{dy}{dx} on one side, we get
\displaystyle \frac{dy}{dx}=\frac{y-x}{x+y}=\frac{\frac{y}{x}-1}{1+\frac{y}{x}}\qquad...(1)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=\frac{v-1}{v+1}
Shifting v on the right-hand side and taking LCM, we get
\displaystyle x\frac{dv}{dx}=\frac{v-1-v(v+1)}{v+1}
Simplifying the numerator, we get
\displaystyle x\frac{dv}{dx}=-\frac{v^2+1}{v+1}
Separating the variables, we get
\displaystyle -\frac{v+1}{v^2+1}\,dv=\frac{dx}{x}
Integrating both sides, we get
\displaystyle -\int\frac{v+1}{v^2+1}\,dv=\int\frac{dx}{x}
Splitting the integral on the left-hand side, we get
\displaystyle -\int\frac{v}{v^2+1}\,dv-\int\frac{dv}{v^2+1}=\int\frac{dx}{x}\qquad...(2)
Using the standard integral formulas
\displaystyle \int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C
and \displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x+C
we get from equation (2)
\displaystyle -\frac12\log(v^2+1)-\tan^{-1}v=\log|x|+C
Substituting \displaystyle v=\frac{y}{x}, we get
\displaystyle -\frac12\log\left(1+\frac{y^2}{x^2}\right)-\tan^{-1}\left(\frac{y}{x}\right)=\log|x|+C
Simplifying the logarithmic term, we get
\displaystyle -\frac12\log(x^2+y^2)+\log|x|-\tan^{-1}\left(\frac{y}{x}\right)=\log|x|+C
Cancelling \displaystyle \log|x| from both sides, we get
\displaystyle -\frac12\log(x^2+y^2)-\tan^{-1}\left(\frac{y}{x}\right)=C\qquad...(3)
Using the condition \displaystyle y=1 when \displaystyle x=1, we get
\displaystyle -\frac12\log2-\frac{\pi}{4}=C
Substituting this value of \displaystyle C in equation (3), we get
\displaystyle -\frac12\log(x^2+y^2)-\tan^{-1}\left(\frac{y}{x}\right)=-\frac12\log2-\frac{\pi}{4}
Multiplying both sides by \displaystyle -2, we get
\displaystyle \log(x^2+y^2)+2\tan^{-1}\left(\frac{y}{x}\right)=\log2+\frac{\pi}{2}
Hence, the required particular solution is
\boxed{\displaystyle \log(x^2+y^2)+2\tan^{-1}\left(\frac{y}{x}\right)=\frac{\pi}{2}+\log2}
Question 12: Differential Equations Exercise 9.4
12. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle x^2\,dy+(xy+y^2)\,dx=0;\; y=1\text{ when }x=1
Solution
Think First… First find the general solution. Then use the given condition to determine the value of the arbitrary constant. Write the equation in the form \displaystyle \frac{dy}{dx}. Since the RHS depends only on \displaystyle \frac{y}{x}, substitute \displaystyle y=vx.
We have,
\displaystyle x^2\,dy+(xy+y^2)\,dx=0
Rewrite the equation to collect \displaystyle \frac{dy}{dx} on one side, we get
\displaystyle \frac{dy}{dx}=-\frac{xy+y^2}{x^2}=-\left(\frac{y}{x}+\frac{y^2}{x^2}\right)\qquad...(1)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=-(v+v^2)
Taking \displaystyle v to the right-hand side, we get
\displaystyle x\frac{dv}{dx}=-(2v+v^2)
Factorising the numerator, we get
\displaystyle x\frac{dv}{dx}=-v(v+2)
Separating the variables, we get
\displaystyle \frac{dv}{v(v+2)}=-\frac{dx}{x}\qquad...(2)
Resolving the left-hand side into partial fractions, we get
\displaystyle \frac{1}{v(v+2)}=\frac12\left(\frac1v-\frac1{v+2}\right)\qquad...(3)
Substituting equation (3) into equation (2) and integrating both sides, we get
\displaystyle \frac12\int\left(\frac1v-\frac1{v+2}\right)\,dv=-\int\frac{dx}{x}
Using the standard integral formula
\displaystyle \int\frac{dx}{x}=\log|x|+C
we get
\displaystyle \frac12\left(\log|v|-\log|v+2|\right)=-\log|x|+\log C
Combining the logarithms, we get
\displaystyle \frac12\log\left|\frac{v}{v+2}\right|=-\log|x|+\log C
Multiplying both sides by \displaystyle 2, we get
\displaystyle \log\left|\frac{v}{v+2}\right|=-2\log|x|+\log C
💡(Maths Better Note: You may wonder why we have written \displaystyle \log C instead of \displaystyle 2\log C after multiplying both sides by \displaystyle 2. Since \displaystyle 2\log C=\log C^2, and \displaystyle C^2 is again an arbitrary positive constant, we simply write it as \displaystyle \log C for simplicity.)
Using log properties on the RHS, we get
\displaystyle \log\left|\frac{v}{v+2}\right|=\log\left(\frac{C}{x^2}\right)
Taking antilogarithm on both sides, we get
\displaystyle \frac{v}{v+2}=\frac{C}{x^2}
Substituting \displaystyle v=\frac{y}{x} in the above equation, we get
\displaystyle \frac{\frac{y}{x}}{\frac{y}{x}+2}=\frac{C}{x^2}
Simplifying the LHS by taking LCM, we get
\displaystyle \frac{y}{y+2x}=\frac{C}{x^2}\qquad...(4)
Using the condition \displaystyle y=1 when \displaystyle x=1 in equation (4), we get
\displaystyle \frac13=C
Substituting this value of \displaystyle C in equation (4), we get
\displaystyle \frac{y}{y+2x}=\frac{1}{3x^2}
Hence, the required particular solution is
\boxed{\displaystyle 3x^2y=y+2x}
Question 13: Differential Equations 9.4
13. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle \left[x\sin^2\left(\frac{y}{x}\right)-y\right]dx+x\,dy=0;\; y=\frac{\pi}{4}\text{ when }x=1
Solution
Think First… Since this is a particular solution, first find the general solution of the differential equation. Then use the given condition to determine the arbitrary constant. Write the equation in the form \displaystyle \frac{dy}{dx}. Since the RHS is a function of \displaystyle \frac{y}{x} only, substitute \displaystyle y=vx.
We have,
\displaystyle \left[x\sin^2\left(\frac{y}{x}\right)-y\right]dx+x\,dy=0
Rewriting the equation in the form \displaystyle \frac{dy}{dx}, we get
\displaystyle \frac{dy}{dx}=\frac{y}{x}-\sin^2\left(\frac{y}{x}\right)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation, we get
\displaystyle v+x\frac{dv}{dx}=v-\sin^2v
Cancelling \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=-\sin^2v
Separating the variables, we get
\displaystyle \cosec^2v\,dv=-\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\cosec^2v\,dv=-\int\frac{dx}{x}\qquad...(1)
Using the standard integral formulas
\displaystyle \int\cosec^2x\,dx=-\cot x+C
and \displaystyle \int\frac{dx}{x}=\log|x|+C
we get from equation (1)
\displaystyle -\cot v=-\log|x|+C
or \displaystyle \cot v=\log|x|+C
Substituting \displaystyle v=\frac{y}{x}, we get
\displaystyle \cot\left(\frac{y}{x}\right)=\log|x|+C\qquad...(2)
Using the condition \displaystyle y=\frac{\pi}{4} when \displaystyle x=1 in equation (2), we get
\displaystyle \cot\frac{\pi}{4}=\log1+C
\displaystyle 1=0+C \Rightarrow C=1
Substituting this value of \displaystyle C in equation (2), we get
\displaystyle \cot\left(\frac{y}{x}\right)=\log|x|+1
Since \displaystyle 1=\log e, we get
\displaystyle \cot\left(\frac{y}{x}\right)=\log|x|+\log e
Using the property of logarithms, we get
\displaystyle \cot\left(\frac{y}{x}\right)=\log|ex|
Hence, the required particular solution is
\boxed{\displaystyle \cot\left(\frac{y}{x}\right)=\log|ex|}
Question 14: Differential Equations
14. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle \frac{dy}{dx}-\frac{y}{x}+\cosec\left(\frac{y}{x}\right)=0;\; y=0\text{ when }x=1
Solution
Think First… Since this is a particular solution, first find the general solution of the differential equation. Then use the given condition to determine the arbitrary constant. Since the RHS is a function of \displaystyle \frac{y}{x} only, substitute \displaystyle y=vx.
We have,
\displaystyle \frac{dy}{dx}-\frac{y}{x}+\cosec\left(\frac{y}{x}\right)=0
Rewriting the equation to have \displaystyle \frac{dy}{dx} on one side, we get
\displaystyle \frac{dy}{dx}=\frac{y}{x}-\cosec\left(\frac{y}{x}\right)\qquad...(1)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=v-\cosec v
or \displaystyle x\frac{dv}{dx}=-\cosec v
Separating the variables, we get
\displaystyle \sin v\,dv=-\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\sin v\,dv=-\int\frac{dx}{x}
\displaystyle \Rightarrow -\cos v=-\log|x|+C
or \displaystyle \cos v=\log|x|+C
Substituting \displaystyle v=\frac{y}{x} in the above equation, we get
\displaystyle \cos\left(\frac{y}{x}\right)=\log|x|+C\qquad...(2)
Using the condition \displaystyle y=0 when \displaystyle x=1 in equation (2), we get
\displaystyle \cos0=\log1+C
\displaystyle 1=0+C \Rightarrow C=1
Substituting this value of \displaystyle C in equation (2), we get
\displaystyle \cos\left(\frac{y}{x}\right)=\log|x|+1
Since \displaystyle 1=\log e, we get
\displaystyle \cos\left(\frac{y}{x}\right)=\log|x|+\log e
Using the property of logarithms, we get
\displaystyle \cos\left(\frac{y}{x}\right)=\log|ex|
Hence, the required particular solution is
\boxed{\displaystyle \cos\left(\frac{y}{x}\right)=\log|ex|}
Question 15: Differential Equations Initial Value Problem
15. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle 2xy+y^2-2x^2\frac{dy}{dx}=0;\; y=2\text{ when }x=1
Solution
Think First… Since this is a particular solution, first find the general solution of the differential equation. Then use the given condition to determine the arbitrary constant. Write the equation in the form \displaystyle \frac{dy}{dx}. Since the RHS is a function of \displaystyle \frac{y}{x} only, substitute \displaystyle y=vx.
We have,
\displaystyle 2xy+y^2-2x^2\frac{dy}{dx}=0
Taking \displaystyle \frac{dy}{dx} on one side, we get
\displaystyle \frac{dy}{dx}=\frac{2xy+y^2}{2x^2}=\frac{y}{x}+\frac12\left(\frac{y}{x}\right)^2\qquad...(1)
Since the RHS is a function of \displaystyle \frac{y}{x} only, the given differential equation is homogeneous.
Substituting \displaystyle y=vx, we have
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
Substituting these values into the differential equation (1), we get
\displaystyle v+x\frac{dv}{dx}=v+\frac12v^2
Cancelling \displaystyle v from both sides, we get
\displaystyle x\frac{dv}{dx}=\frac12v^2
Separating the variables, we get
\displaystyle \frac{2}{v^2}\,dv=\frac{dx}{x}
Integrating both sides, we get
\displaystyle \int\frac{2}{v^2}\,dv=\int\frac{dx}{x}
Using the standard integral formulas, we get
\displaystyle -\frac{2}{v}=\log|x|+C
Substituting \displaystyle v=\frac{y}{x} in the above equation, we get
\displaystyle -\frac{2x}{y}=\log|x|+C\qquad...(2)
Using the condition \displaystyle y=2 when \displaystyle x=1 in equation (2), we get
\displaystyle -1=\log1+C
\displaystyle \Rightarrow C=-1
Substituting this value of \displaystyle C in equation (2), we get
\displaystyle -\frac{2x}{y}=\log|x|-1
or \displaystyle \frac{2x}{y}=1-\log|x|
Rearranging the terms, we get
\displaystyle y=\frac{2x}{1-\log|x|}
Hence, the required particular solution is
\boxed{\displaystyle y=\frac{2x}{1-\log|x|},\quad x\ne0,\;x\ne e}
Maths Better Tip! In Questions 13 and 14, we introduced \displaystyle 1=\log e to combine the logarithms and match the NCERT answer. We could have avoided that step there, just as we have done here by rearranging the equation directly to express \displaystyle y. Both approaches are mathematically correct. Choose the one that gives the required final form.
Question 16: Differential Equations 9.4 – MCQ
16. A homogeneous differential equation of the form
\displaystyle \frac{dx}{dy}=h\!\left(\frac{x}{y}\right) can be solved by making the substitution
- (A) \displaystyle y=vx
- (B) \displaystyle v=yx
- (C) \displaystyle x=vy
- (D) \displaystyle x=v
Solution
Think First… When the differential equation is written in the form \displaystyle \frac{dx}{dy}=h\!\left(\frac{x}{y}\right), the substitution should contain the ratio \displaystyle \frac{x}{y}.
Since the equation involves the ratio \displaystyle \frac{x}{y}, let
\displaystyle \frac{x}{y}=v
or equivalently,
\displaystyle x=vy
✅ Hence, the correct answer is (C).
Question 17: Differential Equations 9.4 – MCQ
17. Which of the following is a homogeneous differential equation?
- (A) \displaystyle (4x+6y+5)\,dy-(3y+2x+4)\,dx=0
- (B) \displaystyle (xy)\,dx-(x^3+y^3)\,dy=0
- (C) \displaystyle (x^3+2y^2)\,dx+2xy\,dy=0
- (D) \displaystyle y^2\,dx+(x^2-xy-y^2)\,dy=0
Solution
Think First… A differential equation of the form \displaystyle M\,dx+N\,dy=0 is homogeneous if both \displaystyle M and \displaystyle N are homogeneous functions of the same degree.
Checking each option:
- (A) Not homogeneous because of the constant terms \displaystyle 5 and \displaystyle 4. Hence, \displaystyle \frac{dy}{dx} cannot be expressed as a function of \displaystyle \frac{y}{x} alone.
- (B) Not homogeneous since \displaystyle xy is of degree \displaystyle 2, whereas \displaystyle x^3+y^3 is of degree \displaystyle 3. Hence, \displaystyle \frac{dy}{dx} cannot be expressed as a function of \displaystyle \frac{y}{x} alone.
- (C) Not homogeneous since \displaystyle x^3 is of degree \displaystyle 3, while \displaystyle 2y^2 is of degree \displaystyle 2. Hence, \displaystyle \frac{dy}{dx} cannot be expressed as a function of \displaystyle \frac{y}{x} alone.
- (D) Here, \displaystyle y^2 and \displaystyle x^2-xy-y^2 are both homogeneous functions of degree \displaystyle 2. Writing the equation in the form \displaystyle \frac{dy}{dx}, we get
\displaystyle \frac{dy}{dx}=\frac{y^2}{x^2-xy-y^2}=\frac{\left(\frac{y}{x}\right)^2}{1-\frac{y}{x}-\left(\frac{y}{x}\right)^2}
Thus, \displaystyle \frac{dy}{dx} is a function of \displaystyle \frac{y}{x} alone. Hence, the given differential equation is homogeneous.
✅ Hence, the correct answer is (D).
Maths Better Tip! You can often eliminate options quickly without checking every term. If an equation contains constant terms like \displaystyle 4,\;5, it is usually not homogeneous. Also, every term in a homogeneous function must have the same degree. If the terms have different degrees, you can rule out that option immediately.
Common Mistakes to Avoid
- Not identifying a homogeneous differential equation: Before solving, check whether the right-hand side can be written as a function of \displaystyle \frac{y}{x} or \displaystyle \frac{x}{y}.
- Using the wrong substitution: If the equation is in terms of \displaystyle \frac{y}{x}, use \displaystyle y=vx. If it is in terms of \displaystyle \frac{x}{y}, use \displaystyle x=vy.
- Differentiating the substitution incorrectly: Remember that if \displaystyle y=vx, then \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}. Similarly, if \displaystyle x=vy, then \displaystyle \frac{dx}{dy}=v+y\frac{dv}{dy}.
- Not separating the variables correctly: After substitution, simplify the equation first and then separate the variables before integrating.
- Making mistakes in partial fractions: Decompose the rational expression correctly before integrating whenever required.
- Forgetting the constant of integration: Always add the constant of integration on one side after integrating both sides of the differential equation.
- Not applying the initial condition correctly: First obtain the general solution and then use the given point to determine the constant of integration.
- Ignoring the required final form: Rearrange or simplify the final answer whenever necessary to match the form given in NCERT.
- Making mistakes while handling logarithms: Use logarithmic properties carefully, and introduce \displaystyle \log C whenever it gives a cleaner solution.
- Rushing the algebra after substitution: Most errors in homogeneous differential equations occur after substituting \displaystyle y=vx or \displaystyle x=vy, rather than during integration.
Continue Learning
In Exercise 9.4 we covered the Homogeneous Differential Equations. In this method, the given differential equation is first converted into a variable separable form by using a suitable substitution. This exercise also includes questions involving general as well as particular solutions.
Before starting Exercise 9.5, revise the following concepts:
- How to identify a homogeneous differential equation by checking whether the right-hand side is a function of \displaystyle \frac{y}{x} or \displaystyle \frac{x}{y} only.
- The correct substitution: use \displaystyle y=vx when the equation is in terms of \displaystyle \frac{y}{x}, and \displaystyle x=vy when it is in terms of \displaystyle \frac{x}{y}.
- How to differentiate the substitution correctly, such as \displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx} or \displaystyle \frac{dx}{dy}=v+y\frac{dv}{dy}.
- How to simplify the equation after substitution and then separate the variables before integrating.
- Standard integration formulas and the method of partial fractions, wherever required.
- How to apply the given initial condition to obtain the particular solution after finding the general solution.
- Logarithmic properties such as \displaystyle \log a+\log b=\log(ab) and \displaystyle \log a-\log b=\log\left(\frac ab\right) for simplifying the final answer.
- The importance of rearranging the final answer, whenever required, to match the form given in NCERT.
Explore More
A good understanding of the Homogeneous Differential Equations method makes solving many first-order differential equations much easier. The key is to first identify whether the given equation is homogeneous, choose the correct substitution such as \displaystyle y=vx or \displaystyle x=vy, convert it into a variable separable form and then integrate carefully. If an initial condition is given, first obtain the general solution and then determine the constant of integration. Finally, simplify or rearrange the result, whenever required, to match the NCERT form. With regular practice, you will quickly recognize homogeneous differential equations and solve them confidently using this powerful technique.
All the best and keep learning 👍
