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Differential Equations question

2022 · 29 Jul · Shift 1 · Q31
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  5. /2022 · 29 Jul · Shift 1 · Q31

Differential Equations question

2022 · 29 Jul · Shift 1 · Q31

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let the solution curve y=y(x)y=y(x)y=y(x) of the differential equation (1+e2x)(dy dx+y)=1\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1(1+e2x)( dxdy​+y)=1 pass through the point (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​). Then, lim⁡x→∞exy(x)\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)x→∞lim​exy(x) is equal to :
  1. A
    π4\frac{\pi}{4}4π​
  2. B
    3π4\frac{3\pi}{4}43π​
  3. C
    π2\frac{\pi}{2}2π​
  4. D
    3π2\frac{3\pi}{2}23π​
View written solutionFree

Correct answer: B

  1. Rewrite the differential equation

Given

(1+e2x)(dydx+y)=1(1+e^{2x})\left(\frac{dy}{dx}+y\right)=1(1+e2x)(dxdy​+y)=1

so

dydx+y=11+e2x.\frac{dy}{dx}+y=\frac{1}{1+e^{2x}}.dxdy​+y=1+e2x1​.

This is a linear first-order differential equation.

  1. Use the integrating factor

The integrating factor is

IF=e∫1 dx=ex.IF=e^{\int 1\,dx}=e^x.IF=e∫1dx=ex.

Multiplying the equation by exe^xex:

exdydx+exy=ex1+e2x.e^x\frac{dy}{dx}+e^x y=\frac{e^x}{1+e^{2x}}.exdxdy​+exy=1+e2xex​.

The left side becomes

ddx(exy)=ex1+e2x.\frac{d}{dx}(e^x y)=\frac{e^x}{1+e^{2x}}.dxd​(exy)=1+e2xex​.
  1. Integrate both sides
exy=∫ex1+e2x dx+C.e^x y=\int \frac{e^x}{1+e^{2x}}\,dx + C.exy=∫1+e2xex​dx+C.

Let

t=ex  ⟹  dt=exdx.t=e^x \implies dt=e^x dx.t=ex⟹dt=exdx.

Then

∫ex1+e2x dx=∫dt1+t2=tan⁡−1(t)=tan⁡−1(ex).\int \frac{e^x}{1+e^{2x}}\,dx=\int \frac{dt}{1+t^2}=\tan^{-1}(t)=\tan^{-1}(e^x).∫1+e2xex​dx=∫1+t2dt​=tan−1(t)=tan−1(ex).

Hence,

exy=tan⁡−1(ex)+C.e^x y=\tan^{-1}(e^x)+C.exy=tan−1(ex)+C.

So,

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

Given that the curve passes through

(0,π2).\left(0,\frac{\pi}{2}\right).(0,2π​).

Substitute x=0x=0x=0, y=π2y=\frac{\pi}{2}y=2π​:

e0⋅π2=tan⁡−1(e0)+C.e^0\cdot \frac{\pi}{2}=\tan^{-1}(e^0)+C.e0⋅2π​=tan−1(e0)+C.

Since e0=1e^0=1e0=1 and tan⁡−1(1)=π4\tan^{-1}(1)=\frac{\pi}{4}tan−1(1)=4π​,

π2=π4+C\frac{\pi}{2}=\frac{\pi}{4}+C2π​=4π​+C

which gives

C=π4.C=\frac{\pi}{4}.C=4π​.

Thus,

exy=tan⁡−1(ex)+π4.e^x y=\tan^{-1}(e^x)+\frac{\pi}{4}.exy=tan−1(ex)+4π​.
  1. Evaluate the limit

We need

lim⁡x→∞exy(x).\lim_{x\to\infty} e^x y(x).x→∞lim​exy(x).

From the expression above,

lim⁡x→∞exy(x)=lim⁡x→∞(tan⁡−1(ex)+π4).\lim_{x\to\infty} e^x y(x)=\lim_{x\to\infty}\left(\tan^{-1}(e^x)+\frac{\pi}{4}\right).x→∞lim​exy(x)=x→∞lim​(tan−1(ex)+4π​).

As x→∞x\to\inftyx→∞, ex→∞e^x\to\inftyex→∞, so

tan⁡−1(ex)→π2.\tan^{-1}(e^x)\to \frac{\pi}{2}.tan−1(ex)→2π​.

Therefore,

lim⁡x→∞exy(x)=π2+π4=3π4.\lim_{x\to\infty} e^x y(x)=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}.x→∞lim​exy(x)=2π​+4π​=43π​.
  1. Match with the options
3π4\frac{3\pi}{4}43π​

corresponds to Option B.

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