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Differential Equations question

2022 · 29 Jun · Shift 2 · Q29
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  5. /2022 · 29 Jun · Shift 2 · Q29

Differential Equations question

2022 · 29 Jun · Shift 2 · Q29

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If y = y(x) is the solution of the differential equation (1+e2x)dydx+2(1+y2)ex=0\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0(1+e2x)dxdy​+2(1+y2)ex=0 and y (0) = 0, then 6(y′(0)+(y(log⁡e3))2)6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)6(y′(0)+(y(loge​3​))2) is equal to
  1. A
    2
  2. B
    −-− 2
  3. C
    −-− 4
  4. D
    −-− 1
View written solutionFree

Correct answer: C

  1. Given differential equation
(1+e2x)dydx+2(1+y2)ex=0(1+e^{2x})\frac{dy}{dx}+2(1+y^2)e^x=0(1+e2x)dxdy​+2(1+y2)ex=0

with initial condition

y(0)=0.y(0)=0.y(0)=0.

We need to find

6(y′(0)+(y(ln⁡3))2).6\left(y'(0)+\left(y\left(\ln\sqrt{3}\right)\right)^2\right).6(y′(0)+(y(ln3​))2).
  1. Rewrite the differential equation

Divide both sides by 1+e2x1+e^{2x}1+e2x:

dydx=−2ex(1+y2)1+e2x.\frac{dy}{dx}=-\frac{2e^x(1+y^2)}{1+e^{2x}}.dxdy​=−1+e2x2ex(1+y2)​.

Now separate variables:

dy1+y2=−2ex1+e2xdx.\frac{dy}{1+y^2}=-\frac{2e^x}{1+e^{2x}}dx.1+y2dy​=−1+e2x2ex​dx.
  1. Integrate both sides

We know

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

For the right side, let t=ext=e^xt=ex, so dt=exdxdt=e^x dxdt=exdx:

∫−2ex1+e2xdxn=−2∫dt1+t2=−2tan⁡−1t+C=−2tan⁡−1(ex)+C.\int -\frac{2e^x}{1+e^{2x}}dx n= -2\int \frac{dt}{1+t^2} = -2\tan^{-1}t + C = -2\tan^{-1}(e^x)+C.∫−1+e2x2ex​dxn=−2∫1+t2dt​=−2tan−1t+C=−2tan−1(ex)+C.

Hence,

tan⁡−1y=−2tan⁡−1(ex)+C.\tan^{-1}y=-2\tan^{-1}(e^x)+C.tan−1y=−2tan−1(ex)+C.
  1. Use the initial condition

At x=0x=0x=0, y=0y=0y=0:

tan⁡−1(0)=−2tan⁡−1(1)+C.\tan^{-1}(0)=-2\tan^{-1}(1)+C.tan−1(0)=−2tan−1(1)+C.

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

0=−2⋅π4+C=−π2+C.0=-2\cdot \frac{\pi}{4}+C=-\frac{\pi}{2}+C.0=−2⋅4π​+C=−2π​+C.

So,

C=π2.C=\frac{\pi}{2}.C=2π​.

Thus,

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

So,

y=tan⁡(π2−2tan⁡−1(ex))=cot⁡(2tan⁡−1(ex)).y=\tan\left(\frac{\pi}{2}-2\tan^{-1}(e^x)\right)=\cot\left(2\tan^{-1}(e^x)\right).y=tan(2π​−2tan−1(ex))=cot(2tan−1(ex)).

Using

cot⁡(2θ)=1−tan⁡2θ2tan⁡θ,\cot(2\theta)=\frac{1-\tan^2\theta}{2\tan\theta},cot(2θ)=2tanθ1−tan2θ​,

with tan⁡θ=ex\tan\theta=e^xtanθ=ex, we get

y=1−e2x2ex.y=\frac{1-e^{2x}}{2e^x}.y=2ex1−e2x​.

This simplifies to

y=−ex−e−x2=−sinh⁡x.y=-\frac{e^x-e^{-x}}{2}=-\sinh x.y=−2ex−e−x​=−sinhx.
  1. Find y′(0)y'(0)y′(0)

Since

y=−sinh⁡x,y=-\sinh x,y=−sinhx,

we have

y′=−cosh⁡x.y'=-\cosh x.y′=−coshx.

Therefore,

y′(0)=−cosh⁡0=−1.y'(0)=-\cosh 0=-1.y′(0)=−cosh0=−1.
  1. Find y(ln⁡3)y\left(\ln\sqrt{3}\right)y(ln3​)

Let

x=ln⁡3.x=\ln\sqrt{3}.x=ln3​.

Then

ex=3,e−x=13.e^x=\sqrt{3}, \qquad e^{-x}=\frac{1}{\sqrt{3}}.ex=3​,e−x=3​1​.

So

y(ln⁡3)=1−e2x2ex=1−323=−223=−13.y\left(\ln\sqrt{3}\right)=\frac{1-e^{2x}}{2e^x} =\frac{1-3}{2\sqrt{3}} =\frac{-2}{2\sqrt{3}} =-\frac{1}{\sqrt{3}}.y(ln3​)=2ex1−e2x​=23​1−3​=23​−2​=−3​1​.

Hence,

(y(ln⁡3))2=13.\left(y\left(\ln\sqrt{3}\right)\right)^2=\frac{1}{3}.(y(ln3​))2=31​.
  1. Compute the required expression
6(y′(0)+(y(ln⁡3))2)=6(−1+13)=6(−23)=−4.6\left(y'(0)+\left(y\left(\ln\sqrt{3}\right)\right)^2\right) =6\left(-1+\frac{1}{3}\right) =6\left(-\frac{2}{3}\right) =-4.6(y′(0)+(y(ln3​))2)=6(−1+31​)=6(−32​)=−4.
  1. Check options

The value is

−4.-4.−4.

So the correct option is C.

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