Differential Equations: Chapter 9 Links
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Differential Equations Exercise 9.5 introduces First Order Linear Differential Equations and the Integrating Factor (I.F.) method used to solve them. In this exercise, you will learn how to write a differential equation in the standard linear form, find its integrating factor, and use it to obtain the required solution.
\displaystyle \frac{dy}{dx}+Py=Q,\qquad \text{I.F.}=e^{\int P\,dx},\qquad y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
The main idea is to convert the given differential equation into the standard linear form, calculate the integrating factor, and then apply a fixed formula to obtain the solution. The same method can also be used when the equation is given in terms of dx/dy.
By practising these questions, you will learn to identify first order linear differential equations, find the correct integrating factor, and solve them confidently. These concepts are important for the CBSE board examinations and CUET.
Key Concepts
Before solving Exercise 9.5, it is important to understand the standard form of a First Order Linear Differential Equation, the concept of an Integrating Factor (I.F.), and the steps used to obtain its solution.
1. First Order Linear Differential Equation
A first order linear differential equation can be written in the standard form
\displaystyle \frac{dy}{dx}+Py=Q
where P and Q are constants or functions of x only.
If the equation is given in terms of dx/dy, it is written as
\displaystyle \frac{dx}{dy}+Px=Q
where P and Q are constants or functions of y only.
2. Integrating Factor (I.F.)
For an equation in the form
\displaystyle \frac{dy}{dx}+Py=Q
the Integrating Factor (I.F.) is given by
\displaystyle \text{I.F.}=e^{\int P\,dx}
Similarly, if the equation is in the form
\displaystyle \frac{dx}{dy}+Px=Q
then
\displaystyle \text{I.F.}=e^{\int P\,dy}
3. Solution Formula
After finding the Integrating Factor, the solution is obtained using
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
For equations involving dx/dy, use
\displaystyle x(\text{I.F.})=\int(Q\times\text{I.F.})\,dy+C
4. Steps to Solve
- Write the differential equation in the standard linear form.
- Identify P and Q.
- Find the Integrating Factor (I.F.).
- Apply the solution formula.
- Evaluate the integral and simplify the final answer.
5. When to Integrate with Respect to y
If the differential equation is given in terms of dx/dy, treat y as the independent variable. Find the Integrating Factor using dy and perform all integrations with respect to y.
6. Important Points
- First write the equation in the standard linear form before finding the Integrating Factor.
- The Integrating Factor depends only on P.
- Use the variable with respect to which the differential equation is written while finding the Integrating Factor and integrating.
- Always include the constant of integration C in the final answer.
Quick Reference Table
| Type | Form |
|---|---|
| dy/dx |
First Order Linear DE (FOLDE): P and Q are constants or functions of x. I.F.: Solution: |
| dx/dy |
First Order Linear DE (FOLDE): P and Q are constants or functions of y. I.F.: Solution: |
Note: Throughout this exercise, FOLDE stands for First Order Linear Differential Equation. It is a Maths Better memory aid for quick revision and is not a standard mathematical abbreviation.
Before You Begin… Before finding the Integrating Factor, first write the differential equation in its FOLDE form. Then identify P and Q correctly. The Integrating Factor depends only on P, not on Q.
Let us now solve all the NCERT questions step by step of Differential Equations 9.5.
Question 1: First Order Linear Differential Equations
1. Find the general solution of the following differential equation:
\displaystyle \frac{dy}{dx}+2y=\sin x
Solution
Think First… Write the given equation in the standard FOLDE form \displaystyle \frac{dy}{dx}+Py=Q. Identify P and Q, find the Integrating Factor (I.F.), then multiply the entire equation by the I.F. This converts the left-hand side into the derivative of a product.
We have, \displaystyle \frac{dy}{dx}+2y=\sin x\qquad ...(1)
The given differential equation is already in the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
where P=2 and Q=\sin x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int2\,dx}=e^{2x}
Multiplying both sides of the differential equation (1) by the Integrating Factor, we get
\displaystyle e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}\sin x
The left-hand side is the derivative of \displaystyle ye^{2x} (i.e. \displaystyle y(\text{I.F.})). Therefore, the above equation becomes
\displaystyle \frac{d}{dx}\!\left(ye^{2x}\right)=e^{2x}\sin x
Integrating both sides with respect to x, we get
\displaystyle ye^{2x}=\int e^{2x}\sin x\,dx+C\qquad ...(2)
(Note: Equation (2) gives the solution of a FOLDE in the form \displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C. Once you understand how this result is obtained, you may use this formula directly, as we will do in the remaining questions.)
Let \displaystyle I=\int\underbrace{\sin x}_{\mathrm{I}}\;\underbrace{e^{2x}}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula, (same as Question 21 of Exercise 7.6)
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
Since \displaystyle \int e^{2x}\,dx=\frac12e^{2x} and \displaystyle \frac{d}{dx}(\sin x)=\cos x, we get
\displaystyle I=\frac12e^{2x}\sin x-\frac12\int e^{2x}\cos x\,dx
or \displaystyle I=\frac12e^{2x}\sin x-\frac12I_1\qquad ...(3)
where, \displaystyle I_1=\int e^{2x}\cos x\,dx
Again integrating by parts,
Now, Consider \displaystyle I_1=\int\underbrace{\cos x}_{\mathrm{I}}\;\underbrace{e^{2x}}_{\mathrm{II}}\,dx
We get \displaystyle I_1=\frac12e^{2x}\cos x+\frac12\int e^{2x}\sin x\,dx
Since \displaystyle \int e^{2x}\sin x\,dx=I, we get
\displaystyle I_1=\frac12e^{2x}\cos x+\frac12I
Substituting the value of I_1 in equation (3), we get
\displaystyle I=\frac12e^{2x}\sin x-\frac12\left(\frac12e^{2x}\cos x+\frac12I\right)
Simplifying, we get
\displaystyle I=\frac{2}{5}e^{2x}\sin x-\frac15e^{2x}\cos x
or \displaystyle I=\frac{e^{2x}}5\left(2\sin x-\cos x\right)
Substituting this value in equation (2), we get
\displaystyle ye^{2x}=\frac{e^{2x}}5\left(2\sin x-\cos x\right)+C
Dividing both sides by \displaystyle e^{2x},
\displaystyle y=\frac{2\sin x-\cos x}{5}+Ce^{-2x}
Hence, the required general solution is
\boxed{\displaystyle y=\frac15\left(2\sin x-\cos x\right)+Ce^{-2x}}
Question 2: Differential Equations 9.5
2. Find the general solution of the following differential equation:
\displaystyle \frac{dy}{dx}+3y=e^{-2x}
Solution
Think First… Write the given equation in the standard FOLDE form \displaystyle \frac{dy}{dx}+Py=Q. Identify P and Q, find the Integrating Factor (I.F.) and then use the shortcut solution formula introduced in Question 1.
We have, \displaystyle \frac{dy}{dx}+3y=e^{-2x}\qquad ...(1)
The given differential equation is already in the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
where P=3 and Q=e^{-2x}.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int3\,dx}=e^{3x}
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (1) is
\displaystyle ye^{3x}=\int e^{-2x}\cdot e^{3x}\,dx+C
\displaystyle =\int e^x\,dx+C=e^x+C
Dividing both sides by \displaystyle e^{3x}, we get
\displaystyle y=e^{-2x}+Ce^{-3x}
Hence, the required general solution is
\boxed{\displaystyle y=e^{-2x}+Ce^{-3x}}
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 9.5 introduces the method of solving First Order Linear Differential Equations using the Integrating Factor (I.F.). With sufficient practice, you will learn to write a differential equation in its standard form, find the Integrating Factor and apply the solution formula confidently. This is one of the most important techniques for solving first-order differential equations. Now, let’s proceed to the next question.
Question 3: Differential Equations
3. Find the general solution of the following differential equation:
\displaystyle \frac{dy}{dx}+\frac{y}{x}=x^2
Solution
Think First… Write the given equation in the standard FOLDE form \displaystyle \frac{dy}{dx}+Py=Q. Identify P and Q, find the Integrating Factor (I.F.) and then use the shortcut solution formula.
We have, \displaystyle \frac{dy}{dx}+\frac{y}{x}=x^2\qquad ...(1)
The given differential equation is already in the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
where \displaystyle P=\frac1x and \displaystyle Q=x^2.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac1x\,dx}=e^{\log x}=x
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (1) is
\displaystyle xy=\int x^2\cdot x\,dx+C
\displaystyle =\int x^3\,dx+C=\frac{x^4}{4}+C
Hence, the required general solution is
\boxed{\displaystyle xy=\frac{x^4}{4}+C}
Question 4: Differential Equations 9.5
4. Find the general solution of the following differential equation:
\displaystyle \frac{dy}{dx}+(\sec x)y=\tan x,\; \left(0\le x\lt\frac{\pi}{2}\right)
Solution
Think First… Write the given equation in the standard FOLDE form \displaystyle \frac{dy}{dx}+Py=Q. Identify P and Q, find the Integrating Factor (I.F.) and then use the shortcut solution formula.
We have, \displaystyle \frac{dy}{dx}+(\sec x)y=\tan x\qquad ...(1)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\sec x and \displaystyle Q=\tan x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\sec x\,dx}=e^{\log(\sec x+\tan x)}=\sec x+\tan x
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (1) is
\displaystyle y(\sec x+\tan x)=\int\tan x(\sec x+\tan x)\,dx+C
\displaystyle =\int(\sec x\tan x+\tan^2x)\,dx+C
Using \displaystyle \tan^2x=\sec^2x-1, we get
\displaystyle y(\sec x+\tan x)=\int(\sec x\tan x+\sec^2x-1)\,dx+C
\displaystyle =\sec x+\tan x-x+C
Hence, the required general solution is
\boxed{\displaystyle y(\sec x+\tan x)=\sec x+\tan x-x+C}
Maths Better Tip… Before finding the Integrating Factor (I.F.), always write the differential equation in the standard \displaystyle \frac{dy}{dx}+Py=Q form. This means the coefficient of \displaystyle \frac{dy}{dx} must be 1. If it is not, first divide the entire equation by that coefficient and then proceed to find the I.F.
Question 5: Differential Equations 9.5
5. Find the general solution of the following differential equation:
\displaystyle \cos^2x\frac{dy}{dx}+y=\tan x,\; \left(0\le x<\frac{\pi}{2}\right)
Solution
Think First… The given equation is not in the standard FOLDE form. First divide both sides by the coefficient of \displaystyle \frac{dy}{dx}, then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle \cos^2x\frac{dy}{dx}+y=\tan x\qquad ...(1)
Dividing both sides of equation (1) by \displaystyle \cos^2x, we get
\displaystyle \frac{dy}{dx}+\sec^2x\,y=\tan x\sec^2x\qquad ...(2)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\sec^2x and \displaystyle Q=\tan x\sec^2x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\sec^2x\,dx}=e^{\tan x}
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (2) is
\displaystyle ye^{\tan x}=\int\tan x\sec^2x\,e^{\tan x}\,dx+C
Let \displaystyle t=\tan x. Then, \displaystyle dt=\sec^2x\,dx.
Therefore, \displaystyle ye^{\tan x}=\int te^t\,dt+C\qquad ...(3)
Applying the Integration by Parts formula, we get
\displaystyle I=\int \underbrace{t}_{\mathrm{I}}\;\underbrace{e^t}_{\mathrm{II}}\,dt=e^t(t-1)+C
Substituting \displaystyle t=\tan x, equation (3) becomes
\displaystyle ye^{\tan x}=e^{\tan x}(\tan x-1)+C
Dividing both sides by \displaystyle e^{\tan x}, we get
\displaystyle y=\tan x-1+Ce^{-\tan x}
Hence, the required general solution is
\boxed{\displaystyle y=(\tan x-1)+Ce^{-\tan x}}
Question 6: Exercise 9.5
6. Find the general solution of the following differential equation:
\displaystyle x\frac{dy}{dx}+2y=x^2\log x
Solution
Think First… The given equation is not in the standard FOLDE form. First divide both sides by the coefficient of \displaystyle \frac{dy}{dx}, then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle x\frac{dy}{dx}+2y=x^2\log x\qquad ...(1)
Dividing both sides of equation (1) by x, we get
\displaystyle \frac{dy}{dx}+\frac{2}{x}y=x\log x\qquad ...(2)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\frac2x and \displaystyle Q=x\log x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac2x\,dx}=e^{2\log|x|}=x^2
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (2) is
\displaystyle x^2y=\int x\log x\cdot x^2\,dx+C
\displaystyle =\int x^3\log x\,dx+C\qquad ...(3)
Let \displaystyle I=\int \underbrace{\log x}_{\mathrm{I}}\;\underbrace{x^3}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
We get, \displaystyle I=\log x\left(\frac{x^4}{4}\right)-\int\frac1x\left(\frac{x^4}{4}\right)dx
\displaystyle =\frac{x^4}{4}\log x-\frac14\int x^3\,dx
\displaystyle =\frac{x^4}{4}\log x-\frac{x^4}{16}
Substituting this value in equation (3), we get
\displaystyle x^2y=\frac{x^4}{4}\log x-\frac{x^4}{16}+C
Dividing both sides by x^2, we get
\displaystyle y=\frac{x^2}{4}\log x-\frac{x^2}{16}+Cx^{-2}
\displaystyle =\frac{x^2}{16}(4\log x-1)+Cx^{-2}
Hence, the required general solution is
\boxed{\displaystyle y=\frac{x^2}{16}(4\log x-1)+Cx^{-2}}
Question 7: Differential Equations 9.5
7. Find the general solution of the following differential equation:
\displaystyle x\log x\,\frac{dy}{dx}+y=\frac{2\log x}{x}
Solution
Think First… The given equation is not in the standard FOLDE form. First divide both sides by the coefficient of \displaystyle \frac{dy}{dx}, then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle x\log x\,\frac{dy}{dx}+y=\frac{2\log x}{x}\qquad ...(1)
Dividing both sides of equation (1) by \displaystyle x\log x, we get
\displaystyle \frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x^2}\qquad ...(2)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\frac1{x\log x} and \displaystyle Q=\frac2{x^2}.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac1{x\log x}\,dx}
To evaluate \displaystyle \int\frac1{x\log x}\,dx, let \displaystyle t=\log x. Then, \displaystyle dt=\frac1x\,dx. Hence,
\displaystyle \int\frac1{x\log x}\,dx=\int\frac1t\,dt=\log t=\log(\log x)
\displaystyle \therefore \text{I.F.}=e^{\log(\log x)}=\log x\; (\because e^{\log N}=N)
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (2) is
\displaystyle y\log x=\int\frac{2\log x}{x^{2}}\,dx+C\qquad ...(3)
Let \displaystyle I=2\int \underbrace{\log x}_{\mathrm{I}}\;\underbrace{x^{-2}}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
We get, \displaystyle I=2\left[\log x\left(-\frac1x\right)-\int\frac1x\left(-\frac1x\right)dx\right]
\displaystyle =-\frac{2\log x}{x}+2\int\frac1{x^2}\,dx
\displaystyle =-\frac{2\log x}{x}-\frac2x
Substituting this value in equation (3), we get
\displaystyle y\log x=-\frac{2\log x}{x}-\frac2x+C
Hence, the required general solution is
\boxed{\displaystyle y\log x=-\frac{2}{x}\left(1+\log 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 9.5
8. Find the general solution of the following differential equation:
\displaystyle (1+x^2)\,dy+2xy\,dx=\cot x\,dx,\; (x\ne0)
Solution
Think First… The given equation is not in the standard FOLDE form. First rewrite it in terms of \displaystyle \frac{dy}{dx}, then divide by the coefficient of \displaystyle \frac{dy}{dx}, identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle (1+x^2)\,dy+2xy\,dx=\cot x\,dx\qquad ...(1)
Dividing equation (1) by dx, we get
\displaystyle (1+x^2)\frac{dy}{dx}+2xy=\cot x\qquad ...(2)
Now dividing both sides of equation (2) by \displaystyle 1+x^2, we get
\displaystyle \frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{\cot x}{1+x^2}\qquad ...(3)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\frac{2x}{1+x^2} and \displaystyle Q=\frac{\cot x}{1+x^2}.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac{2x}{1+x^2}\,dx}
To evaluate \displaystyle \int\frac{2x}{1+x^2}\,dx, let \displaystyle t=1+x^2. Then, \displaystyle dt=2x\,dx. Hence,
\displaystyle \int\frac{2x}{1+x^2}\,dx=\int\frac1t\,dt=\log t=\log(1+x^2)
\displaystyle \therefore \text{I.F.}=e^{\log(1+x^2)}=1+x^2\; (\because e^{\log N}=N)
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (3) is
\displaystyle y(1+x^2)=\int\frac{\cot x}{1+x^2}(1+x^2)\,dx+C
\displaystyle =\int\cot x\,dx+C=\log|\sin x|+C
Therefore, \displaystyle (1+x^2)y=\log|\sin x|+C
Dividing both sides by \displaystyle 1+x^2, we get
\displaystyle y=\frac{\log|\sin x|+C}{1+x^2}
Hence, the required general solution is
\boxed{\displaystyle y=(1+x^2)^{-1}\log|\sin x|+C(1+x^2)^{-1}}
Question 9: Differential Equations 9.5
9. Find the general solution of the following differential equation:
\displaystyle x\frac{dy}{dx}+y-x+xy\cot x=0,\; (x\ne0)
Solution
Think First… The coefficient of \displaystyle \frac{dy}{dx} is not 1. First divide the entire equation by this coefficient to write it in the standard FOLDE form. Then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle x\frac{dy}{dx}+y-x+xy\cot x=0\qquad ...(1)
Rearranging equation (1), we get
\displaystyle x\frac{dy}{dx}+y+xy\cot x=x\qquad ...(2)
Now dividing both sides of equation (2) by \displaystyle x, we get
\displaystyle \frac{dy}{dx}+\left(\frac1x+\cot x\right)y=1\qquad ...(3)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\frac1x+\cot x and \displaystyle Q=1.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\left(\frac1x+\cot x\right)\,dx}
Using \displaystyle \int\frac1x\,dx=\log|x| and \displaystyle \int\cot x\,dx=\log|\sin x|, we get
\displaystyle \int\left(\frac1x+\cot x\right)\,dx=\log|x|+\log|\sin x|=\log|x\sin x|
\displaystyle \therefore \text{I.F.}=e^{\log|x\sin x|}=x\sin x\; (\because e^{\log N}=N)
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the solution of equation (3) is
\displaystyle xy\sin x=\int x\sin x\,dx+C\qquad ...(4)
Let \displaystyle I=\int\underbrace{\sin x}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
Using Integration by Parts,
\displaystyle I=x(-\cos x)-\int(-\cos x)\,dx
\displaystyle =-x\cos x+\sin x
Substituting in the equation (4) above, we obtain
\displaystyle xy\sin x=-x\cos x+\sin x+C
Dividing both sides by \displaystyle x\sin x, we get
\displaystyle y=\frac{-x\cos x+\sin x+C}{x\sin x}
\displaystyle =\frac1x-\cot x+\frac{C}{x\sin x}
Hence, the required general solution is
\boxed{\displaystyle y=\frac1x-\cot x+\frac{C}{x\sin x}}
Question 10: Differential Equations Exercise 9.5
10. Find the general solution of the following differential equation:
\displaystyle (x+y)\frac{dy}{dx}=1
Solution
Think First… The given equation is not in the standard FOLDE form with y as the dependent variable. First rewrite the equation in terms of \displaystyle \frac{dx}{dy}. Then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle (x+y)\frac{dy}{dx}=1\qquad ...(1)
Taking the reciprocal of both sides, we get
\displaystyle \frac{dx}{dy}=x+y\qquad ...(2)
Rearranging equation (2), we get
\displaystyle \frac{dx}{dy}-x=y\qquad ...(3)
Comparing with the standard linear form
\displaystyle \frac{dx}{dy}+Px=Q
we get \displaystyle P=-1 and \displaystyle Q=y.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int(-1)\,dy}=e^{-y}
Using the solution formula
\displaystyle x(\text{I.F.})=\int(Q\times\text{I.F.})\,dy+C
the solution of equation (3) is
\displaystyle xe^{-y}=\int ye^{-y}\,dy+C\qquad ...(4)
Let \displaystyle I=\int\underbrace{y}_{\mathrm{I}}\;\underbrace{e^{-y}}_{\mathrm{II}}\,dy
Using Integration by Parts,
We get, \displaystyle I=y(-e^{-y})-\int(-e^{-y})\,dy
\displaystyle =-ye^{-y}-e^{-y}
\displaystyle =-(y+1)e^{-y}
Substituting in equation (4),
\displaystyle xe^{-y}=-(y+1)e^{-y}+C
Multiplying both sides by \displaystyle e^y, we get
\displaystyle x=-(y+1)+Ce^y
Hence, the required general solution is
\boxed{\displaystyle x+y+1=Ce^y}
Maths Better Tip! There are two types of First Order Linear Differential Equations (FOLDE). Before finding the Integrating Factor (I.F.), first identify which variable the equation is linear in.
- If the equation is linear in y, write it in the form \displaystyle \frac{dy}{dx}+Py=Q and use \displaystyle \text{I.F.}=e^{\int P\,dx}.
- If the equation is linear in x, rewrite it in the form \displaystyle \frac{dx}{dy}+Px=Q and use \displaystyle \text{I.F.}=e^{\int P\,dy}.
Always decide the dependent variable first. Once the equation is written in the correct standard form, the remaining steps—finding the Integrating Factor and applying the shortcut solution formula—are exactly the same.
Question 11: General Solution
11. Find the general solution of the following differential equation:
\displaystyle y\,dx+(x-y^2)\,dy=0
Solution
Think First… The given equation is not in the standard FOLDE form with y as the dependent variable. First rewrite the equation in terms of \displaystyle \frac{dx}{dy}. Then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle y\,dx+(x-y^2)\,dy=0\qquad ...(1)
Dividing equation (1) by dy, we get
\displaystyle y\frac{dx}{dy}+x-y^2=0\qquad ...(2)
Now dividing both sides of equation (2) by \displaystyle y, we get
\displaystyle \frac{dx}{dy}+\frac1yx=y\qquad ...(3)
Comparing with the standard linear form
\displaystyle \frac{dx}{dy}+Px=Q
we get \displaystyle P=\frac1y and \displaystyle Q=y.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int\frac1y\,dy}=e^{\log|y|}=y
(\displaystyle \because e^{\log N}=N)
Using the solution formula
\displaystyle x(\text{I.F.})=\int(Q\times\text{I.F.})\,dy+C
the solution of equation (3) is
\displaystyle xy=\int y\cdot y\,dy+C
\displaystyle =\int y^2\,dy+C=\frac{y^3}{3}+C
Dividing both sides by \displaystyle y, we get
\displaystyle x=\frac{y^2}{3}+\frac{C}{y}
Hence, the required general solution is
\boxed{\displaystyle x=\frac{y^2}{3}+\frac{C}{y}}
Maths Better Tip! There is no fixed rule to decide whether a differential equation should be solved using \displaystyle \frac{dy}{dx} or \displaystyle \frac{dx}{dy}. As a habit, first try writing it in the standard form using \displaystyle \frac{dy}{dx}. If it does not become a First Order Linear Differential Equation, rewrite it in terms of \displaystyle \frac{dx}{dy}.
For example, in Question 12,
\displaystyle (x+3y^2)\frac{dy}{dx}=y\;\Longrightarrow\;\frac{dy}{dx}=\frac{y}{x+3y^2}
This cannot be written in the standard linear form \displaystyle \frac{dy}{dx}+Py=Q. Now write it in terms of \displaystyle \frac{dx}{dy}:
\displaystyle \frac{dx}{dy}=\frac{x+3y^2}{y}=\frac{x}{y}+3y\;\Longrightarrow\;\frac{dx}{dy}-\frac1y\,x=3y
Now it is in the standard linear form and can be solved using the Integrating Factor (I.F.) method.
Question 12: Differential Equations Exercise 9.5
12. Find the general solution of the following differential equation:
\displaystyle (x+3y^2)\frac{dy}{dx}=y,\; (y>0)
Solution
Think First… The given equation is not in the standard FOLDE form with y as the dependent variable. First rewrite the equation in terms of \displaystyle \frac{dx}{dy}. Then identify P and Q, find the Integrating Factor (I.F.) and use the shortcut solution formula.
We have,
\displaystyle (x+3y^2)\frac{dy}{dx}=y\qquad ...(1)
Taking the reciprocal of both sides, we get
\displaystyle \frac{dx}{dy}=\frac{x+3y^2}{y}\qquad ...(2)
Simplifying equation (2), we get
\displaystyle \frac{dx}{dy}-\frac1yx=3y\qquad ...(3)
Comparing with the standard linear form
\displaystyle \frac{dx}{dy}+Px=Q
we get \displaystyle P=-\frac1y and \displaystyle Q=3y.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int-\frac1y\,dy}=e^{-\log y}=\frac1y
(\displaystyle \because e^{-\log N}=\frac1N)
Using the solution formula
\displaystyle x(\text{I.F.})=\int(Q\times\text{I.F.})\,dy+C
the solution of equation (3) is
\displaystyle \frac{x}{y}=\int3\,dy+C
\displaystyle =3y+C
Multiplying both sides by \displaystyle y, we get
\displaystyle x=3y^2+Cy
Hence, the required general solution is
\boxed{\displaystyle x=3y^2+Cy}
Maths Better Tip! So far, we have been finding the general solution, which contains an arbitrary constant C. When an initial condition or a point on the curve is given, first obtain the general solution and then substitute the given values to determine C. The resulting equation is called the particular solution.
Question 13: Particular Solution
13. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle \frac{dy}{dx}+2y\tan x=\sin x,\; y=0\text{ when }x=\frac{\pi}{3}
Solution
Think First… The given equation is already in the standard FOLDE form. Identify P and Q, find the Integrating Factor (I.F.), obtain the general solution and then use the given condition to determine the value of the constant.
We have,
\displaystyle \frac{dy}{dx}+2y\tan x=\sin x\qquad ...(1)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=2\tan x and \displaystyle Q=\sin x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int2\tan x\,dx}
Using \displaystyle \int\tan x\,dx=\log|\sec x|, we get
\displaystyle \int2\tan x\,dx=2\log|\sec x|=\log|\sec^2x|
\displaystyle \therefore \text{I.F.}=e^{\log|\sec^2x|}=\sec^2x\; (\because e^{\log N}=N)
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the general solution of equation (1) is
\displaystyle y\sec^2x=\int\sin x\sec^2x\,dx+C
\displaystyle =\int\sec x\tan x\,dx+C=\sec x+C
Therefore, \displaystyle y=\cos x+C\cos^2x\qquad ...(2)
Using the given condition \displaystyle y=0 when \displaystyle x=\frac{\pi}{3},
We get, \displaystyle 0=\cos\frac{\pi}{3}+C\cos^2\frac{\pi}{3}
\displaystyle 0=\frac12+\frac{C}{4}
\displaystyle C=-2
Substituting this value of C in equation (2), we get
\displaystyle y=\cos x-2\cos^2x
Hence, the required particular solution is
\boxed{\displaystyle y=\cos x-2\cos^2x}
Question 14: Differential Equations
14. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle (1+x^2)\frac{dy}{dx}+2xy=\frac1{1+x^2},\; y=0\text{ when }x=1
Solution
Think First… The coefficient of \displaystyle \frac{dy}{dx} is not 1. First divide the entire equation by this coefficient to write it in the standard FOLDE form. Then identify P and Q, find the Integrating Factor (I.F.), obtain the general solution and then use the given condition to determine the value of the constant.
We have,
\displaystyle (1+x^2)\frac{dy}{dx}+2xy=\frac1{1+x^2}\qquad ...(1)
Dividing both sides of equation (1) by \displaystyle 1+x^2, we get
\displaystyle \frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac1{(1+x^2)^2}\qquad ...(2)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=\frac{2x}{1+x^2} and \displaystyle Q=\frac1{(1+x^2)^2}.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int\frac{2x}{1+x^2}\,dx}
To evaluate \displaystyle \int\frac{2x}{1+x^2}\,dx, let \displaystyle t=1+x^2. Then, \displaystyle dt=2x\,dx. Hence,
\displaystyle \int\frac{2x}{1+x^2}\,dx=\int\frac1t\,dt=\log t=\log(1+x^2)
\displaystyle \therefore \text{I.F.}=e^{\log(1+x^2)}=1+x^2\; (\because e^{\log N}=N)
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the general solution of equation (2) is
\displaystyle y(1+x^2)=\int\frac1{(1+x^2)^2}(1+x^2)\,dx+C
\displaystyle =\int\frac1{1+x^2}\,dx+C=\tan^{-1}x+C
Therefore, \displaystyle y(1+x^2)=\tan^{-1}x+C\qquad ...(3)
Using the given condition \displaystyle y=0 when \displaystyle x=1,
We get, \displaystyle 0=\tan^{-1}(1)+C
\displaystyle 0=\frac{\pi}{4}+C
\displaystyle C=-\frac{\pi}{4}
Substituting this value of C in equation (3), we get
\displaystyle y(1+x^2)=\tan^{-1}x-\frac{\pi}{4}
Hence, the required particular solution is
\boxed{\displaystyle y(1+x^2)=\tan^{-1}x-\frac{\pi}{4}}
Question 15: Differential Equations Initial Value Problem
15. Find the particular solution of the following differential equation satisfying the given condition:
\displaystyle \frac{dy}{dx}-3y\cot x=\sin2x,\; y=2\text{ when }x=\frac{\pi}{2}
Solution
Think First… The given equation is already in the standard FOLDE form. Identify P and Q, find the Integrating Factor (I.F.), obtain the general solution and then use the given condition to determine the value of the constant.
We have,
\displaystyle \frac{dy}{dx}-3y\cot x=\sin2x\qquad ...(1)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=-3\cot x and \displaystyle Q=\sin2x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int-3\cot x\,dx}
Using \displaystyle \int\cot x\,dx=\log|\sin x|, we get
\displaystyle \int-3\cot x\,dx=-3\log|\sin x|=\log|\sin^{-3}x|
Therefore, \displaystyle \text{I.F.}=e^{\log|\sin^{-3}x|}=\sin^{-3}x
(\displaystyle \because e^{\log N}=N)
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the general solution of equation (1) is
\displaystyle y\sin^{-3}x=\int\sin2x\,\sin^{-3}x\,dx+C
Using \displaystyle \sin2x=2\sin x\cos x, we get
\displaystyle y\sin^{-3}x=\int2\cot x\,\cosec x\,dx+C
\displaystyle \therefore y\sin^{-3}x=-2\cosec x+C\qquad ...(2)
Using the given condition \displaystyle y=2 when \displaystyle x=\frac{\pi}{2}, we get
\displaystyle 2=-2+C
\displaystyle C=4
Substituting this value of C in equation (2), we get
\displaystyle y\sin^{-3}x=-2\cosec x+4
Multiplying both sides by \displaystyle \sin^3x,
\displaystyle y=-2\sin^2x+4\sin^3x
Hence, the required particular solution is
\boxed{\displaystyle y=4\sin^3x-2\sin^2x}
Maths Better Tip! In word problems, the biggest challenge is not solving the differential equation—it is forming it correctly. Read the statement carefully, identify what is represented by x, y and \displaystyle \frac{dy}{dx}, then convert the given information into a differential equation. Once the equation is in the standard linear form, the remaining steps are exactly the same as before.
Question 16: Word Problem
16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point \displaystyle (x,y) is equal to the sum of the coordinates of the point.
Solution
Think First… First convert the given statement into a differential equation. Then solve the resulting FOLDE and finally use the given condition to determine the value of the constant.
We know that the slope of the tangent to a curve at any point is \displaystyle \frac{dy}{dx}. Hence, we first express the given information in terms of \displaystyle \frac{dy}{dx} to obtain the required differential equation.
From the given condition, the slope of the tangent is equal to the sum of the coordinates of the point.
Therefore, the differential equation is
\displaystyle \frac{dy}{dx}=x+y
Rearranging, we get
\displaystyle \frac{dy}{dx}-y=x\qquad ...(1)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=-1 and \displaystyle Q=x.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-1)\,dx}=e^{-x}
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the general solution of equation (1) is
\displaystyle ye^{-x}=\int xe^{-x}\,dx+C\qquad ...(2)
Let \displaystyle I=\int\underbrace{x}_{\mathrm{I}}\;\underbrace{e^{-x}}_{\mathrm{II}}\,dx
Using Integration by Parts,
We get, \displaystyle I=x(-e^{-x})-\int(-e^{-x})\,dx
\displaystyle =-xe^{-x}-e^{-x}
\displaystyle =-(x+1)e^{-x}
Substituting in equation (2), we get
\displaystyle ye^{-x}=-(x+1)e^{-x}+C
Multiplying both sides by \displaystyle e^x,
\displaystyle y=-(x+1)+Ce^x\qquad ...(3)
Since the curve passes through the origin, \displaystyle (0,0),
\displaystyle 0=-(0+1)+Ce^0
\displaystyle C=1
Substituting this value of C in equation (3), we get
\displaystyle y=e^x-x-1
Hence, the required equation of the curve is
\boxed{\displaystyle x+y+1=e^x}
Question 17: Differential Equations 9.5
17. Find the equation of a curve passing through the point \displaystyle (0,2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Solution
Think First… First convert the given statement into a differential equation. Then solve the resulting FOLDE and finally use the given condition to determine the value of the constant.
We know that the slope of the tangent to a curve at any point is \displaystyle \frac{dy}{dx}. Hence, we first express the given information in terms of \displaystyle \frac{dy}{dx} to obtain the required differential equation.
From the given condition, the sum of the coordinates exceeds the slope of the tangent by 5.
Therefore,
\displaystyle x+y=\frac{dy}{dx}+5
or,
\displaystyle \frac{dy}{dx}-y=x-5\qquad ...(1)
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=-1 and \displaystyle Q=x-5.
So, the Integrating Factor is
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int(-1)\,dx}=e^{-x}
Using the solution formula
\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C
the general solution of equation (1) is
\displaystyle ye^{-x}=\int(x-5)e^{-x}\,dx+C\qquad ...(2)
Let \displaystyle I=\int\underbrace{(x-5)}_{\mathrm{I}}\;\underbrace{e^{-x}}_{\mathrm{II}}\,dx
Using Integration by Parts,
We get, \displaystyle I=(x-5)(-e^{-x})-\int(-e^{-x})\,dx
\displaystyle =-(x-5)e^{-x}-e^{-x}
\displaystyle =-(x-4)e^{-x}
Substituting in equation (2), we get
\displaystyle ye^{-x}=-(x-4)e^{-x}+C
Multiplying both sides by \displaystyle e^x,
\displaystyle y=-x+4+Ce^x\qquad ...(3)
Since the curve passes through \displaystyle (0,2),
\displaystyle 2=-(0)+4+Ce^0
\displaystyle C=-2
Substituting this value of C in equation (3), we get
\displaystyle y=4-x-2e^x
Hence, the required equation of the curve is
\boxed{\displaystyle y=4-x-2e^x}
Question 18: Differential Equations 9.5 – MCQ
18. The Integrating Factor of the differential equation \displaystyle x\frac{dy}{dx}-y=2x^2 is
- (A) \displaystyle e^{-x}
- (B) \displaystyle e^{-y}
- (C) \displaystyle \frac1x
- (D) \displaystyle x
Solution
Think First… First write the differential equation in the standard FOLDE form \displaystyle \frac{dy}{dx}+Py=Q. Then use the formula \displaystyle \text{I.F.}=e^{\int P\,dx}.
Given,
\displaystyle x\frac{dy}{dx}-y=2x^2
Dividing both sides by \displaystyle x, we get
\displaystyle \frac{dy}{dx}-\frac1xy=2x
Comparing with the standard linear form
\displaystyle \frac{dy}{dx}+Py=Q
we get \displaystyle P=-\frac1x.
Therefore,
\displaystyle \text{I.F.}=e^{\int-\frac1x\,dx}=e^{-\log|x|}=\frac1x
(\displaystyle \because e^{-\log N}=\frac1N)
✅ Hence, the correct answer is (C).
Question 19: Differential Equations 9.5 – MCQ
19. The Integrating Factor of the differential equation \displaystyle (1-y^2)\frac{dx}{dy}+yx=ay,\;(-1<y<1) is
- (A) \displaystyle \frac1{y^2-1}
- (B) \displaystyle \frac1{\sqrt{y^2-1}}
- (C) \displaystyle \frac1{1-y^2}
- (D) \displaystyle \frac1{\sqrt{1-y^2}}
Solution
Think First… First write the differential equation in the standard FOLDE form \displaystyle \frac{dx}{dy}+Px=Q. Then use the formula \displaystyle \text{I.F.}=e^{\int P\,dy}.
Given,
\displaystyle (1-y^2)\frac{dx}{dy}+yx=ay
Dividing both sides by \displaystyle 1-y^2, we get
\displaystyle \frac{dx}{dy}+\frac{y}{1-y^2}x=\frac{ay}{1-y^2}
Comparing with the standard linear form
\displaystyle \frac{dx}{dy}+Px=Q
we get \displaystyle P=\frac{y}{1-y^2}.
Therefore,
\displaystyle \text{I.F.}=e^{\int\frac{y}{1-y^2}\,dy}
To evaluate \displaystyle \int\frac{y}{1-y^2}\,dy, let \displaystyle t=1-y^2. Then, \displaystyle dt=-2y\,dy. Hence,
\displaystyle \int\frac{y}{1-y^2}\,dy=-\frac12\int\frac1t\,dt=-\frac12\log t=-\frac12\log(1-y^2)
Therefore,
\displaystyle \text{I.F.}=e^{-\frac12\log(1-y^2)}=\frac1{\sqrt{1-y^2}}
(\displaystyle \because e^{-\log N^{1/2}}=\frac1{\sqrt N})
✅ Hence, the correct answer is (D).
Maths Better Tip! Whenever you see an integral of the form \displaystyle \int \frac{f'(x)}{f(x)}\,dx, try to identify it directly instead of using substitution. In many MCQs, you can obtain the numerator by simply multiplying and dividing by a suitable constant.
For example,
\displaystyle \int \frac{y}{1-y^2}\,dy=-\frac12\int\frac{-2y}{1-y^2}\,dy=-\frac12\log(1-y^2)+C
This simple observation can save valuable time in both board exams and CUET.
Common Mistakes to Avoid
- Not writing the equation in the standard form: Before finding the Integrating Factor (I.F.), always write the equation in the form \displaystyle \frac{dy}{dx}+Py=Q. If the coefficient of \displaystyle \frac{dy}{dx} is not 1, divide the entire equation by that coefficient first.
- Using the wrong dependent variable: Some questions become linear only after writing the equation in terms of \displaystyle \frac{dx}{dy}. Always identify whether the equation is linear in y or in x.
- Finding the Integrating Factor incorrectly: Always identify \displaystyle P correctly and use \displaystyle \text{I.F.}=e^{\int P\,dx} (or \displaystyle e^{\int P\,dy} when solving in terms of \displaystyle \frac{dx}{dy}).
- Making mistakes while evaluating \displaystyle \int P\,dx: Use the correct substitution or standard integral whenever required before finding the Integrating Factor.
- Using the shortcut formula incorrectly: After finding the Integrating Factor, apply \displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C carefully. When solving in terms of \displaystyle \frac{dx}{dy}, replace y by x accordingly.
- Making mistakes in Integration by Parts: Choose the first function using the ILATE rule and differentiate and integrate the chosen functions carefully.
- Forgetting the constant of integration: Always add the constant of integration after evaluating the integral.
- Not applying the given condition correctly: First obtain the general solution and then use the given point or condition to determine the value of the constant.
- Ignoring the required final form: Rearrange or simplify the final answer, whenever necessary, to match the form given in NCERT.
- Making errors while simplifying the final answer: Use logarithmic, exponential and trigonometric identities carefully before writing the final result.
Continue Learning
In Exercise 9.5, we learn how to solve a First Order Linear Differential Equation (FOLDE) using the Integrating Factor (I.F.) method. This exercise includes finding the Integrating Factor, solving linear differential equations, determining particular solutions using the given conditions and a few multiple-choice questions.
Before starting Misc. Exercise, revise the following concepts:
- How to write a differential equation in the standard form \displaystyle \frac{dy}{dx}+Py=Q (or \displaystyle \frac{dx}{dy}+Px=Q, whenever required).
- How to identify the functions \displaystyle P and \displaystyle Q correctly before finding the Integrating Factor.
- The formula for the Integrating Factor: \displaystyle \text{I.F.}=e^{\int P\,dx} (or \displaystyle e^{\int P\,dy} when solving in terms of \displaystyle \frac{dx}{dy}).
- Standard integration formulas, substitution and Integration by Parts (ILATE rule), wherever required.
- How to use the shortcut solution formula \displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C (or the corresponding formula in terms of x).
- How to apply the given condition after obtaining the general solution to determine the value of the constant.
- Properties of logarithms and exponentials, such as \displaystyle e^{\log N}=N, while simplifying the Integrating Factor and the final answer.
- The importance of simplifying or rearranging the final answer, whenever required, to match the form given in NCERT.
Explore More
A good understanding of the First Order Linear Differential Equation (FOLDE) method makes solving many differential equations much easier. The key is to first write the given equation in the standard form, identify the functions \displaystyle P and \displaystyle Q, find the Integrating Factor (I.F.) and then apply the shortcut solution formula carefully. If the equation is not linear in y, check whether it becomes linear in x by writing it in terms of \displaystyle \frac{dx}{dy}. 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 first order linear differential equations and solve them confidently using the Integrating Factor method.
All the best and keep learning 👍
