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Differential Equations question

2024 · 8 Apr · Shift 1 · Q39
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  5. /2024 · 8 Apr · Shift 1 · Q39

Differential Equations question

2024 · 8 Apr · Shift 1 · Q39

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1+y2)etan⁡xdx+cos⁡2x(1+e2tan⁡x)dy=0,y(0)=1(1+y^2) e^{\tan x} d x+\cos ^2 x(1+e^{2 \tan x}) d y=0, y(0)=1(1+y2)etanxdx+cos2x(1+e2tanx)dy=0,y(0)=1. Then y(π4)y\left(\frac{\pi}{4}\right)y(4π​) is equal to
  1. A
    1e2\frac{1}{e^2}e21​
  2. B
    2e2\frac{2}{e^2}e22​
  3. C
    2e\frac{2}{e}e2​
  4. D
    1e\frac{1}{e}e1​
View written solutionFree

Correct answer: D

  1. Write the differential equation in separable form

Given

(1+y2)etan⁡x dx+cos⁡2x(1+e2tan⁡x) dy=0.(1+y^2)e^{\tan x}\,dx+\cos^2x(1+e^{2\tan x})\,dy=0.(1+y2)etanxdx+cos2x(1+e2tanx)dy=0.

Rearrange:

cos⁡2x(1+e2tan⁡x) dy=−(1+y2)etan⁡x dx.\cos^2x(1+e^{2\tan x})\,dy=-(1+y^2)e^{\tan x}\,dx.cos2x(1+e2tanx)dy=−(1+y2)etanxdx.

So,

dy1+y2=−etan⁡xcos⁡2x(1+e2tan⁡x) dx.\frac{dy}{1+y^2}=-\frac{e^{\tan x}}{\cos^2x(1+e^{2\tan x})}\,dx.1+y2dy​=−cos2x(1+e2tanx)etanx​dx.
  1. Simplify the right-hand side

Use

1+e2tan⁡x=etan⁡x(e−tan⁡x+etan⁡x).1+e^{2\tan x}=e^{\tan x}\left(e^{-\tan x}+e^{\tan x}\right).1+e2tanx=etanx(e−tanx+etanx).

But the best substitution is

t=tan⁡x⇒dt=sec⁡2x dx=dxcos⁡2x.t=\tan x \quad \Rightarrow \quad dt=\sec^2x\,dx=\frac{dx}{\cos^2x}.t=tanx⇒dt=sec2xdx=cos2xdx​.

Thus,

etan⁡xcos⁡2x(1+e2tan⁡x) dx;=et1+e2t dt.\frac{e^{\tan x}}{\cos^2x(1+e^{2\tan x})}\,dx ;= \frac{e^t}{1+e^{2t}}\,dt.cos2x(1+e2tanx)etanx​dx;=1+e2tet​dt.

Hence the equation becomes

dy1+y2=−et1+e2t dt.\frac{dy}{1+y^2}=-\frac{e^t}{1+e^{2t}}\,dt.1+y2dy​=−1+e2tet​dt.
  1. Integrate both sides

Left side:

∫dy1+y2=tan⁡−1y.\int \frac{dy}{1+y^2}=\tan^{-1}y.∫1+y2dy​=tan−1y.

Right side:

∫et1+e2t dt.\int \frac{e^t}{1+e^{2t}}\,dt.∫1+e2tet​dt.

Let

u=et⇒dν=etdt.u=e^t \Rightarrow d\nu=e^t dt.u=et⇒dν=etdt.

Then

∫et1+e2t dt=∫dν1+ν2=tan⁡−1(ν)=tan⁡−1(et).\int \frac{e^t}{1+e^{2t}}\,dt=\int \frac{d\nu}{1+\nu^2}=\tan^{-1}(\nu)=\tan^{-1}(e^t).∫1+e2tet​dt=∫1+ν2dν​=tan−1(ν)=tan−1(et).

Therefore,

tan⁡−1y=−tan⁡−1(et)+C.\tan^{-1}y=-\tan^{-1}(e^t)+C.tan−1y=−tan−1(et)+C.

Since t=tan⁡xt=\tan xt=tanx,

tan⁡−1y=−tan⁡−1(etan⁡x)+C.\tan^{-1}y=-\tan^{-1}(e^{\tan x})+C.tan−1y=−tan−1(etanx)+C.
  1. Use the initial condition y(0)=1y(0)=1y(0)=1

At x=0x=0x=0,

tan⁡0=0,etan⁡0=1,y=1.\tan 0=0, \quad e^{\tan 0}=1, \quad y=1.tan0=0,etan0=1,y=1.

So,

tan⁡−1(1)=−tan⁡−1(1)+C.\tan^{-1}(1)=-\tan^{-1}(1)+C.tan−1(1)=−tan−1(1)+C.

Since tan⁡−1(1)=π4\tan^{-1}(1)=\frac{\pi}{4}tan−1(1)=4π​,

π4=−π4+C⇒C=π2.\frac{\pi}{4}=-\frac{\pi}{4}+C \quad \Rightarrow \quad C=\frac{\pi}{2}.4π​=−4π​+C⇒C=2π​.

Thus,

tan⁡−1y=π2−tan⁡−1(etan⁡x).\tan^{-1}y=\frac{\pi}{2}-\tan^{-1}(e^{\tan x}).tan−1y=2π​−tan−1(etanx).

Using

π2−tan⁡−1u=tan⁡−1(1u)(u>0),\frac{\pi}{2}-\tan^{-1}u=\tan^{-1}\left(\frac{1}{u}\right) \quad (u>0),2π​−tan−1u=tan−1(u1​)(u>0),

we get

tan⁡−1y=tan⁡−1(e−tan⁡x).\tan^{-1}y=\tan^{-1}(e^{-\tan x}).tan−1y=tan−1(e−tanx).

Hence,

y=e−tan⁡x.y=e^{-\tan x}.y=e−tanx.
  1. Compute y(π4)y\left(\frac{\pi}{4}\right)y(4π​)

Since

tan⁡(π4)=1,\tan\left(\frac{\pi}{4}\right)=1,tan(4π​)=1,

we get

y(π4)=e−1=1e.y\left(\frac{\pi}{4}\right)=e^{-1}=\frac{1}{e}.y(4π​)=e−1=e1​.
  1. Compare with options

The correct option is

1e\boxed{\frac{1}{e}}e1​​

which is Option D.

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