Let y=y(x) be the solution of the differential equation \sec ^2 x d x+(e^2 y \tan ^2 x+\tan x) d y=0,0\lt x\lt \frac\pi2, y(\pi / 4)=0 . If y(\pi / 6)=\alpha , then e^8 \alpha is equal to \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q52 with Solution

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Correct answer: 9

Given differential equation

\sec^2 x\,dx+\left(e^{2y}\tan^2 x+\tan x\right)dy=0

with

0<x<\frac{\pi}{2}, \qquad y\left(\frac{\pi}{4}\right)=0

We need to find $\alpha=y\left(\frac{\pi}{6}\right)$ and then compute $e^{8\alpha}$ .

Rewrite using $x$ as a function of $y$

Divide the equation by $dy$ :

\sec^2 x\frac{dx}{dy}+e^{2y}\tan^2 x+\tan x=0

Let

t=\tan x

Then

\frac{dt}{dy}=\sec^2 x\frac{dx}{dy}

So the differential equation becomes

\frac{dt}{dy}+e^{2y}t^2+t=0

That is,

\frac{dt}{dy}+t=-e^{2y}t^2

This is a Bernoulli-type equation.

Use substitution $u=\dfrac{1}{t}$

Let

u=\frac{1}{t}

Then

\frac{d\nu}{dy}=-\frac{1}{t^2}\frac{dt}{dy}

From

\frac{dt}{dy}+t=-e^{2y}t^2

we get

\frac{dt}{dy}=-t-e^{2y}t^2

Hence

\frac{d\nu}{dy}=-\frac{1}{t^2}(-t-e^{2y}t^2)=\frac{1}{t}+e^{2y}=\nu+e^{2y}

\frac{d\nu}{dy}-\nu=e^{2y}

Solve the linear differential equation

The integrating factor is

IF=e^{-y}

Thus,

e^{-y}\frac{d\nu}{dy}-e^{-y}\nu=e^y

which gives

\frac{d}{dy}(\nu e^{-y})=e^y

Integrating,

\nu e^{-y}=e^y+C

Therefore,

\nu=e^{2y}+Ce^y

Since $\nu=\dfrac{1}{t}=\dfrac{1}{\tan x}=\cot x$ , we have

\cot x=e^{2y}+Ce^y

Use the initial condition

Given

y\left(\frac{\pi}{4}\right)=0

At $x=\frac{\pi}{4}$ ,

\cot\frac{\pi}{4}=1

and at $y=0$ ,

e^{2y}=1, \qquad e^y=1

1=1+C

which gives

C=0

Hence the relation simplifies to

\cot x=e^{2y}

Find $\alpha=y\left(\frac{\pi}{6}\right)$

x=\frac{\pi}{6}

we have

\cot\frac{\pi}{6}=\sqrt{3}

e^{2\alpha}=\sqrt{3}

Now square both sides:

e^{4\alpha}=3

Again square:

e^{8\alpha}=9

Final answer

\boxed{9}

The derived answer matches the stored correct answer.

More from Differential Equations

View written solutionFree

Correct answer: 9

Given differential equation

\sec^2 x\,dx+\left(e^{2y}\tan^2 x+\tan x\right)dy=0

with

0<x<\frac{\pi}{2}, \qquad y\left(\frac{\pi}{4}\right)=0

We need to find $\alpha=y\left(\frac{\pi}{6}\right)$ and then compute $e^{8\alpha}$ .

Rewrite using $x$ as a function of $y$

Divide the equation by $dy$ :

\sec^2 x\frac{dx}{dy}+e^{2y}\tan^2 x+\tan x=0

Let

t=\tan x

Then

\frac{dt}{dy}=\sec^2 x\frac{dx}{dy}

So the differential equation becomes

\frac{dt}{dy}+e^{2y}t^2+t=0

That is,

\frac{dt}{dy}+t=-e^{2y}t^2

This is a Bernoulli-type equation.

Use substitution $u=\dfrac{1}{t}$

Let

u=\frac{1}{t}

Then

\frac{d\nu}{dy}=-\frac{1}{t^2}\frac{dt}{dy}

From

\frac{dt}{dy}+t=-e^{2y}t^2

we get

\frac{dt}{dy}=-t-e^{2y}t^2

Hence

\frac{d\nu}{dy}=-\frac{1}{t^2}(-t-e^{2y}t^2)=\frac{1}{t}+e^{2y}=\nu+e^{2y}

\frac{d\nu}{dy}-\nu=e^{2y}

Solve the linear differential equation

The integrating factor is

IF=e^{-y}

Thus,

e^{-y}\frac{d\nu}{dy}-e^{-y}\nu=e^y

which gives

\frac{d}{dy}(\nu e^{-y})=e^y

Integrating,

\nu e^{-y}=e^y+C

Therefore,

\nu=e^{2y}+Ce^y

Since $\nu=\dfrac{1}{t}=\dfrac{1}{\tan x}=\cot x$ , we have

\cot x=e^{2y}+Ce^y

Use the initial condition

Given

y\left(\frac{\pi}{4}\right)=0

At $x=\frac{\pi}{4}$ ,

\cot\frac{\pi}{4}=1

and at $y=0$ ,

e^{2y}=1, \qquad e^y=1

1=1+C

which gives

C=0

Hence the relation simplifies to

\cot x=e^{2y}

Find $\alpha=y\left(\frac{\pi}{6}\right)$

x=\frac{\pi}{6}

we have

\cot\frac{\pi}{6}=\sqrt{3}

e^{2\alpha}=\sqrt{3}

Now square both sides:

e^{4\alpha}=3

Again square:

e^{8\alpha}=9

Final answer

\boxed{9}

The derived answer matches the stored correct answer.

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