Differential Equations Miscellaneous NCERT Solutions

1(i). Determine the order and degree (if defined) of the following differential equation:

\displaystyle \frac{d^2y}{dx^2}+5x\left(\frac{dy}{dx}\right)^2-6y=\log x

Solution

Think First… To find the order, identify the highest order derivative present in the equation. To find the degree, first ensure that the differential equation is a polynomial in its derivatives. Then, the degree is the power of the highest order derivative.

The highest order derivative present is \displaystyle \frac{d^2y}{dx^2}.

Therefore,

\boxed{\text{Order}=2}

The given differential equation is a polynomial in its derivatives. The highest order derivative \displaystyle \frac{d^2y}{dx^2} is raised to the first power.

Therefore,

\boxed{\text{Degree}=1}

1(ii). Determine the order and degree (if defined) of the following differential equation:

\displaystyle \left(\frac{dy}{dx}\right)^3-4\left(\frac{dy}{dx}\right)^2+7y=\sin x

Solution

Think First… To find the order, identify the highest order derivative present in the equation. To find the degree, first ensure that the differential equation is a polynomial in its derivatives. Then, the degree is the power of the highest order derivative.

The highest order derivative present is \displaystyle \frac{dy}{dx}.

Therefore,

\boxed{\text{Order}=1}

The given differential equation is a polynomial in its derivatives. The highest order derivative \displaystyle \frac{dy}{dx} is raised to the third power.

Therefore,

\boxed{\text{Degree}=3}

1(iii). Determine the order and degree (if defined) of the following differential equation:

\displaystyle \frac{d^4y}{dx^4}-\sin\left(\frac{d^3y}{dx^3}\right)=0

Solution

Think First… To find the order, identify the highest order derivative present in the equation. To find the degree, first ensure that the differential equation is a polynomial in its derivatives. If any derivative appears inside a trigonometric, exponential or logarithmic function, the degree is not defined.

The highest order derivative present is \displaystyle \frac{d^4y}{dx^4}.

Therefore,

\boxed{\text{Order}=4}

Now observe that the third order derivative \displaystyle \frac{d^3y}{dx^3} appears inside the trigonometric function \sin\left(\frac{d^3y}{dx^3}\right).

Hence, the given differential equation is not a polynomial in its derivatives.

Therefore, the degree of the differential equation is not defined.

\boxed{\text{Degree is not defined}}

2(i). Verify that the given function is a solution of the corresponding differential equation:

Given,

\displaystyle xy=ae^x+be^{-x}+x^2

Differential Equation:

\displaystyle x\frac{d^2y}{dx^2}+2\frac{dy}{dx}-xy+x^2-2=0

Solution

Think First… To verify a solution, differentiate the given function as required, substitute the derivatives into the differential equation, and check whether the left-hand side simplifies to zero.

From the given function,

\displaystyle xy=ae^x+be^{-x}+x^2

Differentiating both sides with respect to x,

\displaystyle x\frac{dy}{dx}+y=ae^x-be^{-x}+2x\qquad\cdots(1)

Differentiating equation (1) again with respect to x,

\displaystyle x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=ae^x+be^{-x}+2\qquad\cdots(2)

Also, from the given function,

\displaystyle ae^x+be^{-x}=xy-x^2\qquad\cdots(3)

Substituting equation (3) into equation (2),

\displaystyle x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=xy-x^2+2

Rearranging,

\displaystyle x\frac{d^2y}{dx^2}+2\frac{dy}{dx}-xy+x^2-2=0

This is exactly the given differential equation.

Therefore, the given function is a solution of the given differential equation.

2(ii). Verify that the given function is a solution of the corresponding differential equation:

Given,

\displaystyle y=e^x(a\cos x+b\sin x)

Differential Equation:

\displaystyle \frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0

Solution

Think First… Differentiate the given function twice, substitute the derivatives into the differential equation, and simplify. If the left-hand side becomes zero, the given function is a solution.

Given, \displaystyle y=e^x(a\cos x+b\sin x)

Differentiating with respect to x using the product rule,

\displaystyle \frac{dy}{dx}=e^x(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x)

or \displaystyle \frac{dy}{dx}=e^x\big[(a+b)\cos x+(b-a)\sin x\big]\qquad\cdots(1)

Differentiating equation (1) again,

\displaystyle \frac{d^2y}{dx^2}=2e^x(b\cos x-a\sin x)\qquad\cdots(2)

Substituting the values of \displaystyle y,\ \frac{dy}{dx} and \displaystyle \frac{d^2y}{dx^2} into the left-hand side of the differential equation,

We have, \displaystyle \text{L.H.S.}=\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y

\displaystyle =2e^x(b\cos x-a\sin x)-2e^x\big[(a+b)\cos x+(b-a)\sin x\big]+2e^x(a\cos x+b\sin x)

\displaystyle =0=\text{R.H.S.}

Hence verified.

Therefore, the given function is a solution of the given differential equation.

2(iii). Verify that the given function is a solution of the corresponding differential equation:

Given,

\displaystyle y=x\sin 3x

Differential Equation:

\displaystyle \frac{d^2y}{dx^2}+9y-6\cos 3x=0

Solution

Think First… Differentiate the given function twice, substitute the derivatives into the differential equation, and simplify. If the left-hand side becomes zero, the given function is a solution.

Given, \displaystyle y=x\sin 3x

Differentiating with respect to x using the product rule,

\displaystyle \frac{dy}{dx}=\sin 3x+3x\cos 3x\qquad\cdots(1)

Differentiating equation (1) again,

\displaystyle \frac{d^2y}{dx^2}=3\cos 3x+\left(3\cos 3x-9x\sin 3x\right)

\displaystyle \frac{d^2y}{dx^2}=6\cos 3x-9x\sin 3x\qquad\cdots(2)

Substituting the values of \displaystyle y and \displaystyle \frac{d^2y}{dx^2} into the left-hand side of the differential equation,

We have, \displaystyle \text{L.H.S.}=\frac{d^2y}{dx^2}+9y-6\cos 3x

\displaystyle =\left(6\cos 3x-9x\sin 3x\right)+9x\sin 3x-6\cos 3x

\displaystyle =0=\text{R.H.S.}

Hence verified.

Therefore, the given function is a solution of the given differential equation.

2(iv). Verify that the given function is a solution of the corresponding differential equation:

Given,

\displaystyle x^2=2y^2\log y

Differential Equation:

\displaystyle (x^2+y^2)\frac{dy}{dx}-xy=0

Solution

Think First… Differentiate the given equation implicitly with respect to x, simplify to obtain \frac{dy}{dx}, and substitute it into the given differential equation. If the left-hand side becomes zero, the given function is a solution.

Given, \displaystyle x^2=2y^2\log y

Differentiating both sides with respect to x,

\displaystyle 2x=2\left(2y\log y+y\right)\frac{dy}{dx}

\displaystyle x=y(2\log y+1)\frac{dy}{dx}\qquad\cdots(1)

From the given equation,

\displaystyle x^2=2y^2\log y

Adding y^2 to both sides,

\displaystyle x^2+y^2=y^2(2\log y+1)\qquad\cdots(2)

Substituting equations (1) and (2) into the left-hand side of the differential equation,

We have, \displaystyle \text{L.H.S.}=(x^2+y^2)\frac{dy}{dx}-xy

\displaystyle =y^2(2\log y+1)\cdot\frac{x}{y(2\log y+1)}-xy

\displaystyle =xy-xy=0=\text{R.H.S.}

Hence verified.

Therefore, the given function is a solution of the given differential equation.

3. Prove that x^2-y^2=C(x^2+y^2)^2 is the general solution of the differential equation:

\displaystyle (x^3-3xy^2)\,dx=(y^3-3x^2y)\,dy

Solution

Think First… To prove that a given function is the general solution, differentiate it implicitly and simplify the result. If the obtained differential equation matches the given one, the result is verified.

Given, \displaystyle x^2-y^2=C(x^2+y^2)^2

Differentiating both sides with respect to x,

\displaystyle 2x-2y\frac{dy}{dx}=C\cdot2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)

Since

\displaystyle C=\frac{x^2-y^2}{(x^2+y^2)^2}

Substituting this value of C in the above equation, we get

\displaystyle 2x-2y\frac{dy}{dx}=\frac{4(x^2-y^2)}{x^2+y^2}\left(x+y\frac{dy}{dx}\right)

Multiplying both sides by \displaystyle \frac{x^2+y^2}{2},

\displaystyle (x-y\frac{dy}{dx})(x^2+y^2)=2(x^2-y^2)\left(x+y\frac{dy}{dx}\right)

Expanding both sides,

\displaystyle x^3+xy^2-x^2y\frac{dy}{dx}-y^3\frac{dy}{dx}=2x^3-2xy^2+2x^2y\frac{dy}{dx}-2y^3\frac{dy}{dx}

Collecting like terms,

\displaystyle x^3-3xy^2=\left(y^3-3x^2y\right)\frac{dy}{dx}

Multiplying both sides by dx,

\displaystyle (x^3-3xy^2)\,dx=(y^3-3x^2y)\,dy

This is exactly the given differential equation.

Therefore, \displaystyle x^2-y^2=C(x^2+y^2)^2 is the general solution of the given differential equation.

4. Find the general solution of the differential equation:

\displaystyle \frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0

Solution

Think First… Rewrite the differential equation so that all terms involving y are on one side and all terms involving x are on the other. Then integrate both sides separately.

Given, \displaystyle \frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0

Rearranging the given differential equation,

\displaystyle \frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}

Separating the variables,

\displaystyle \frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}

Integrating both sides,

\displaystyle \int\frac{dy}{\sqrt{1-y^2}}=-\int\frac{dx}{\sqrt{1-x^2}}

Using the standard formula \displaystyle \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C, we get

\displaystyle \sin^{-1}y=-\sin^{-1}x+C

Therefore, the general solution is

\boxed{\sin^{-1}x+\sin^{-1}y=C}

5. Show that the general solution of the differential equation

\displaystyle \frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0

is given by \displaystyle x+y+1=A(1-x-y-2xy), where A is a parameter.

Solution

Think First… This is a Variable Separable Differential Equation. Separate the variables, integrate both sides, and simplify the result to obtain the required form.

Given, \displaystyle \frac{dy}{dx}=-\frac{y^2+y+1}{x^2+x+1}

Separating the variables,

\displaystyle \frac{dy}{y^2+y+1}=-\frac{dx}{x^2+x+1}

Completing the square in the denominator on both sides, we get

\displaystyle \frac{dy}{\left(y+\frac12\right)^2+\frac34}=-\frac{dx}{\left(x+\frac12\right)^2+\frac34}

or \displaystyle \frac{dy}{\left(y+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}=-\frac{dx}{\left(x+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}

Using the standard formula

\displaystyle \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac xa\right)+C

where \displaystyle a=\frac{\sqrt3}{2}, integrating both sides gives

\displaystyle \frac{2}{\sqrt3}\tan^{-1}\left(\frac{y+\frac12}{\frac{\sqrt3}{2}}\right)=-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{x+\frac12}{\frac{\sqrt3}{2}}\right)+C\qquad ...(1)

Since, \displaystyle \frac{y+\frac12}{\frac{\sqrt3}{2}}=\left(y+\frac12\right)\cdot\frac2{\sqrt3}=\frac{2y+1}{\sqrt3}

and \displaystyle \frac{x+\frac12}{\frac{\sqrt3}{2}}=\left(x+\frac12\right)\cdot\frac2{\sqrt3}=\frac{2x+1}{\sqrt3}

Therefore, equation (1) becomes

\displaystyle \frac{2}{\sqrt3}\tan^{-1}\left(\frac{2y+1}{\sqrt3}\right)=-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2x+1}{\sqrt3}\right)+C

Multiplying throughout by \displaystyle \frac{\sqrt3}{2} and absorbing the constant into a new constant C_1, we get

\displaystyle \tan^{-1}\left(\frac{2y+1}{\sqrt3}\right)+\tan^{-1}\left(\frac{2x+1}{\sqrt3}\right)=C_1

Taking tangent on both sides and using

\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}

We get, \displaystyle \frac{\frac{2y+1}{\sqrt3}+\frac{2x+1}{\sqrt3}}{1-\frac{(2y+1)(2x+1)}3}=\tan C_1

Taking the LCM in the numerator and denominator,

\displaystyle \Rightarrow \frac{3(2x+2y+2)}{\sqrt3(3-(2x+1)(2y+1))}=\tan C_1

\displaystyle \frac{6(x+y+1)}{3-(2x+1)(2y+1)}=\sqrt3\tan C_1

Simplifying further, we get

\displaystyle \frac{(x+y+1)}{(1-x-y-2xy)}=A

where \displaystyle A=\frac{1}{\sqrt3}\tan C_1,

Hence, we obtain

\boxed{\displaystyle x+y+1=A(1-x-y-2xy)}

6. Find the equation of the curve passing through the point \displaystyle \left(0,\frac{\pi}{4}\right) whose differential equation is:

\displaystyle \sin x\cos y\,dx+\cos x\sin y\,dy=0

Solution

Think First… Separate the variables by dividing both sides appropriately. Integrate to obtain the general solution, then use the given point to determine the constant of integration.

Given,

\displaystyle \sin x\cos y\,dx+\cos x\sin y\,dy=0

Dividing both sides by \displaystyle \cos x\cos y, we get

\displaystyle \tan x\,dx+\tan y\,dy=0

Integrating both sides,

\displaystyle \int\tan x\,dx+\int\tan y\,dy=0

\displaystyle -\log|\cos x|-\log|\cos y|=C

or \displaystyle \log|\cos x\cos y|=\log C_1

Taking antilog on both sides,

\displaystyle \cos x\cos y=C_1\qquad ...(1)

Using the given point \displaystyle \left(0,\frac{\pi}{4}\right),

We get, \displaystyle \cos0\cdot\cos\frac{\pi}{4}=C_1

\displaystyle 1\cdot\frac{1}{\sqrt2}=C_1

\displaystyle \Rightarrow C_1=\frac{1}{\sqrt2}

Substituting the value of constant in equation (1), we get

\displaystyle \cos x\cos y=\frac{1}{\sqrt2}

or \displaystyle \cos y=\frac{\sec x}{\sqrt2}

Hence, the required equation of the curve is

\boxed{\cos y=\frac{\sec x}{\sqrt2}}

7. Find the particular solution of the differential equation

\displaystyle (1+e^{2x})\,dy+(1+y^2)e^x\,dx=0,\; \text{given that }y=1\text{ when }x=0

Solution

Think First… Separate the variables by keeping all the terms involving y on one side and all the terms involving x on the other. Integrate both sides and then use the given initial condition to find the constant of integration.

Given,

\displaystyle (1+e^{2x})\,dy+(1+y^2)e^x\,dx=0

Separating the variables,

\displaystyle \frac{dy}{1+y^2}=-\frac{e^x\,dx}{1+e^{2x}}

Integrating both sides,

\displaystyle \int\frac{dy}{1+y^2}=-\int\frac{e^x}{1+e^{2x}}\,dx

\displaystyle \Rightarrow \tan^{-1}y=-\int\frac{e^x}{1+e^{2x}}\,dx\qquad ...(1)

For the integral on the right-hand side,

Let \displaystyle t=e^x\quad\Rightarrow\quad dt=e^x\,dx

Therefore, equation (1) gives,

\displaystyle \tan^{-1}y=-\int\frac{dt}{1+t^2}

Integrating on the right-hand side and putting back \displaystyle t=e^x, we get

\displaystyle \tan^{-1}y=-\tan^{-1}(t)+C=-\tan^{-1}(e^x)+C

or \displaystyle \tan^{-1}y+\tan^{-1}(e^x)=C\qquad ...(2)

Using the given condition y=1 when x=0, we get

\displaystyle \tan^{-1}(1)+\tan^{-1}(1)=C

\displaystyle \frac{\pi}{4}+\frac{\pi}{4}=C \Rightarrow C=\frac{\pi}{2}

Therefore, from equation (2), the required particular solution is

\boxed{\tan^{-1}y+\tan^{-1}(e^x)=\frac{\pi}{2}}

8. Solve the differential equation:

\displaystyle ye^{x/y}\,dx=\left(xe^{x/y}+y^2\right)\,dy,\; (y\ne0)

Solution

Think First… Before solving any differential equation, first identify its type. Sometimes a suitable substitution can transform the equation into a much simpler form, making it easier to solve.

Given, \displaystyle ye^{x/y}\,dx=\left(xe^{x/y}+y^2\right)\,dy

Dividing both sides by \displaystyle ye^{x/y}\,dy, we get

\displaystyle \frac{dx}{dy}-\frac{x}{y}=ye^{-x/y}\qquad ...(1)

Let \displaystyle v=\frac{x}{y}

Then, \displaystyle x=vy\quad\Rightarrow\quad \frac{dx}{dy}=v+y\frac{dv}{dy}

Substituting in equation (1),

\displaystyle v+y\frac{dv}{dy}-v=ye^{-v}

\displaystyle \Rightarrow \frac{dv}{dy}=e^{-v}

Separating the variables,

\displaystyle e^v\,dv=dy

Integrating both sides,

\displaystyle \int e^v\,dv=\int dy

\displaystyle e^v=y+C

Substituting \displaystyle v=\frac{x}{y}, we get

\displaystyle e^{x/y}=y+C

Hence, the required solution is

\boxed{e^{x/y}=y+C}

9. Find a particular solution of the differential equation

\displaystyle (x-y)(dx+dy)=dx-dy,\; \text{given that }y=-1\text{ when }x=0

Solution

Think First… The expression x-y appears repeatedly in the differential equation. Using the substitution t=x-y simplifies the equation into a Variable Separable Differential Equation.

Given, \displaystyle (x-y)(dx+dy)=dx-dy

Using the substitution \displaystyle t=x-y

We get, \displaystyle dt=dx-dy

Also, from \displaystyle dt=dx-dy

We obtain, \displaystyle dy=dx-dt

Substituting these in the given differential equation, we get

\displaystyle t\left(dx+dx-dt\right)=dt

\displaystyle t(2dx-dt)=dt

or \displaystyle 2t\,dx-t\,dt=dt

Taking all the terms involving dt to one side,

\displaystyle 2t\,dx=(t+1)\,dt

Separating the variables,

\displaystyle \frac{t+1}{2t}\,dt=dx

Writing the left-hand side in a simpler form,

\displaystyle \frac12\left(1+\frac1t\right)dt=dx

Integrating both sides,

\displaystyle \frac12\int\left(1+\frac1t\right)dt=\int dx

\displaystyle \frac12\left(t+\log|t|\right)=x+C

Substituting t=x-y, we get

\displaystyle \frac12\left(x-y+\log|x-y|\right)=x+C

Multiplying throughout by 2,

\displaystyle x-y+\log|x-y|=2x+C_1\qquad ...(1)

Using the given condition x=0,\;y=-1,

\displaystyle 0-(-1)+\log|0-(-1)|=C_1

\displaystyle 1+\log1=C_1

Since \displaystyle \log1=0, we get

\displaystyle C_1=1

Substituting \displaystyle C_1=1 in equation (1), we get

\displaystyle x-y+\log|x-y|=2x+1

Subtracting \displaystyle x-y from both sides,

\displaystyle \log|x-y|=2x+1-(x-y)

Simplifying, we get the required particular solution as

\boxed{\displaystyle \log|x-y|=x+y+1}

10. Solve the differential equation:

\displaystyle \left[\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right]\frac{dx}{dy}=1,\; (x\ne0)

Solution

Think First… The equation is given in terms of \displaystyle \frac{dx}{dy}. Rewrite it in terms of \displaystyle \frac{dy}{dx} first. It then becomes a First Order Linear Differential Equation in y.

Given, \displaystyle \left[\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right]\frac{dx}{dy}=1

Taking the reciprocal of both sides, we get

\displaystyle \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}

Rearranging the terms,

\displaystyle \frac{dy}{dx}+\frac{1}{\sqrt{x}}\,y=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}\qquad ...(1)

Comparing with the standard form

\displaystyle \frac{dy}{dx}+Py=Q\qquad ...(2)

We have, \displaystyle P=\frac1{\sqrt{x}},\qquad Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}

The Integrating Factor (I.F.) is

\displaystyle \text{I.F.}=e^{\int\frac{dx}{\sqrt{x}}}=e^{\int x^{-1/2}\,dx}=e^{2\sqrt{x}}

We know the solution of equation type (2) is given by

\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C

Therefore, solution of equation (1) will be

\displaystyle ye^{2\sqrt{x}}=\int\frac{e^{2\sqrt{x}}e^{-2\sqrt{x}}}{\sqrt{x}}\,dx+C

or \displaystyle ye^{2\sqrt{x}}=\int\frac1{\sqrt{x}}\,dx+C

\displaystyle \Rightarrow ye^{2\sqrt{x}}=2\sqrt{x}+C

Hence, the required solution is

\boxed{\displaystyle ye^{2\sqrt{x}}=2\sqrt{x}+C}

11. Find a particular solution of the differential equation

\displaystyle \frac{dy}{dx}+y\cot x=4x\cosec x,\; (x\ne0)

given that y=0 when x=\dfrac{\pi}{2}.

Solution

Think First… This is a First Order Linear Differential Equation. Find the Integrating Factor first, multiply the entire equation by it, integrate both sides and then use the given initial condition to determine the constant of integration.

Given, \displaystyle \frac{dy}{dx}+y\cot x=4x\cosec x\qquad ...(1)

Comparing with the standard form

\displaystyle \frac{dy}{dx}+Py=Q\qquad ...(2)

We get, \displaystyle P=\cot x,\qquad Q=4x\cosec x

The Integrating Factor (I.F.) is

\displaystyle \text{I.F.}=e^{\int\cot x\,dx}=e^{\log|\sin x|}=\sin x

We know the solution of equation type (2) is given by

\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C

Therefore, solution of equation (1) will be

\displaystyle y\sin x=\int4x\sin x\cosec x\,dx+C

or \displaystyle y\sin x=\int4x\,dx+C

\displaystyle \Rightarrow y\sin x=2x^2+C\qquad ...(2)

Using the given condition y=0 when x=\dfrac{\pi}{2},

\displaystyle 0\cdot\sin\frac{\pi}{2}=2\left(\frac{\pi}{2}\right)^2+C

\displaystyle 0=\frac{\pi^2}{2}+C

Hence, \displaystyle C=-\frac{\pi^2}{2}

Substituting this value of C in the equation (2), we get the required solution as

\boxed{\displaystyle y\sin x=2x^2-\frac{\pi^2}{2}\; (x\ne0)}

12. Find a particular solution of the differential equation

\displaystyle (x+1)\frac{dy}{dx}=2e^{-y}-1,\; \text{given that }y=0\text{ when }x=0

Solution

Think First… Separate the variables by keeping all the terms involving y on one side and all the terms involving x on the other. After integrating, use the given initial condition to determine the constant of integration.

We have, \displaystyle (x+1)\frac{dy}{dx}=2e^{-y}-1

Separating the variables, we get

\displaystyle \frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}

Multiplying the numerator and denominator on the left-hand side by e^y,

\displaystyle \frac{e^y\,dy}{2-e^y}=\frac{dx}{x+1}\qquad ...(1)

Using the substitution \displaystyle t=2-e^y

We get, \displaystyle dt=-e^y\,dy

or \displaystyle -dt=e^y\,dy

Substituting these values in equation (1), we get

\displaystyle -\frac{dt}{t}=\frac{dx}{x+1}

Integrating both sides,

\displaystyle -\int\frac{dt}{t}=\int\frac{dx}{x+1}

\displaystyle -\log|t|=\log|x+1|+\log C

Substituting t=2-e^y, we get

\displaystyle -\log|2-e^y|=\log|x+1|+\log C

Rearranging the terms,

\displaystyle \log|2-e^y|+\log|x+1|=-\log C

Using the log properties, we get

\displaystyle \log|(2-e^y)(x+1)|=\log C^{-1}

Taking antilogarithm on both sides,

\displaystyle (2-e^y)(x+1)=\frac1C\qquad ...(3)

Using the given condition x=0,\;y=0,

\displaystyle (2-e^0)(0+1)=\frac1C

\displaystyle (2-1)(1)=\frac1C \Rightarrow C=1

Substituting this value of C in equation (3),

We get, \displaystyle (2-e^y)(x+1)=1

\displaystyle 2-e^y=\frac1{x+1}

\displaystyle \Rightarrow e^y=2-\frac1{x+1}

Taking LCM on the right-hand side,

\displaystyle e^y=\frac{2(x+1)-1}{x+1}

\displaystyle e^y=\frac{2x+1}{x+1}

Taking logarithm on both sides, we get the required solution

\boxed{\displaystyle y=\log\left|\frac{2x+1}{x+1}\right|,\qquad x\ne-1}

13. The general solution of the differential equation \displaystyle \frac{y\,dx-x\,dy}{y}=0 is

• (A) \displaystyle xy=C
• (B) \displaystyle x=Cy^2
• (C) \displaystyle y=Cx
• (D) \displaystyle y=Cx^2

Solution

Think First… Rewrite the differential equation in terms of \displaystyle \frac{dy}{dx}. Then separate the variables and integrate both sides.

Given,

\displaystyle \frac{y\,dx-x\,dy}{y}=0

Multiplying both sides by \displaystyle y, we get

\displaystyle y\,dx-x\,dy=0

Rearranging the terms,

\displaystyle x\,dy=y\,dx

Dividing both sides by \displaystyle xy, we get

\displaystyle \frac{dy}{y}=\frac{dx}{x}

Integrating both sides,

\displaystyle \int\frac{dy}{y}=\int\frac{dx}{x}

\displaystyle \log|y|=\log|x|+\log C

Using the property of logarithms,

\displaystyle \log|y|=\log|Cx|

Taking antilogarithm on both sides,

\displaystyle y=Cx

✅ Hence, the correct answer is (C).

14. The general solution of a differential equation of the type \displaystyle \frac{dx}{dy}+P_1x=Q_1 is

• (A) \displaystyle ye^{\int P_1\,dy}=\int\left(Q_1e^{\int P_1\,dy}\right)dy+C
• (B) \displaystyle ye^{\int P_1\,dx}=\int\left(Q_1e^{\int P_1\,dx}\right)dx+C
• (C) \displaystyle xe^{\int P_1\,dy}=\int\left(Q_1e^{\int P_1\,dy}\right)dy+C
• (D) \displaystyle xe^{\int P_1\,dx}=\int\left(Q_1e^{\int P_1\,dx}\right)dx+C

Solution

Think First… The given differential equation is a First Order Linear Differential Equation in x with y as the independent variable. Use the standard solution formula for the equation \displaystyle \frac{dx}{dy}+Px=Q.

Given, \displaystyle \frac{dx}{dy}+P_1x=Q_1\qquad ...(1)

Comparing with the standard linear form

\displaystyle \frac{dx}{dy}+Px=Q\qquad ...(2)

we get \displaystyle P=P_1 and \displaystyle Q=Q_1.

Therefore, the Integrating Factor (I.F.) is

\displaystyle \text{I.F.}=e^{\int P\,dy}=e^{\int P_1\,dy}

We know the solution of equation type (2) is given by

\displaystyle x(\text{I.F.})=\int(Q\times\text{I.F.})\,dy+C

Hence, the general solution of equation (1) is

\displaystyle xe^{\int P_1\,dy}=\int\left(Q_1e^{\int P_1\,dy}\right)dy+C

✅ Hence, the correct answer is (C).

15. The general solution of the differential equation \displaystyle e^x\,dy+\left(ye^x+2x\right)dx=0 is

• (A) \displaystyle xe^y+x^2=C
• (B) \displaystyle xe^y+y^2=C
• (C) \displaystyle ye^x+x^2=C
• (D) \displaystyle ye^y+x^2=C

Solution

Think First… Rewrite the equation in the standard First Order Linear Differential Equation (FOLDE) form. Then find the Integrating Factor and compare the obtained solution with the given options.

Given, \displaystyle e^x\,dy+\left(ye^x+2x\right)dx=0

Dividing both sides by \displaystyle e^x\,dx, we get

\displaystyle \frac{dy}{dx}+y=-2xe^{-x}\qquad ...(1)

Comparing with the standard linear form

\displaystyle \frac{dy}{dx}+Py=Q\qquad ...(2)

We have, \displaystyle P=1,\qquad Q=-2xe^{-x}

Therefore, the Integrating Factor (I.F.) is

\displaystyle \text{I.F.}=e^{\int1\,dx}=e^x

We know the solution of equation type (2) is given by

\displaystyle y(\text{I.F.})=\int(Q\times\text{I.F.})\,dx+C

Thetefore, solution of given equation (1) will be

\displaystyle ye^x=\int(-2xe^{-x}e^x)\,dx+C

or \displaystyle ye^x=\int(-2x)\,dx+C

\displaystyle \Rightarrow ye^x=-x^2+C

or \displaystyle ye^x+x^2=C

✅ Hence, the correct answer is (C).