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Differential Equations question

2021 · 18 Mar · Shift 2 · Q23
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  5. /2021 · 18 Mar · Shift 2 · Q23

Differential Equations question

2021 · 18 Mar · Shift 2 · Q23

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation dydx=(y+1)((y+1)ex2/2−x){{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)dxdy​=(y+1)((y+1)ex2/2−x), 0 < x < 2.1, with y(2) = 0. Then the value of dydx{{dy} \over {dx}}dxdy​ at x = 1 is equal to :
  1. A
    e5/2(1+e2)2{{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}(1+e2)2e5/2​
  2. B
    5e1/2(e2+1)2{{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}(e2+1)25e1/2​
  3. C
    −2e2(1+e2)2- {{2{e^2}} \over {{{(1 + {e^2})}^2}}}−(1+e2)22e2​
  4. D
    −e3/2(e2+1)2{{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}(e2+1)2−e3/2​
View written solutionFree

Correct answer: D

  1. Given differential equation
dydx=(y+1)((y+1)ex2/2−x),0<x<2.1,\frac{dy}{dx}=(y+1)\left((y+1)e^{x^2/2}-x\right), \qquad 0<x<2.1,dxdy​=(y+1)((y+1)ex2/2−x),0<x<2.1,

with

y(2)=0.y(2)=0.y(2)=0.

We need to find dydx∣x=1\left.\dfrac{dy}{dx}\right|_{x=1}dxdy​​x=1​.

  1. Substitute a convenient variable

Let

z=y+1.z=y+1.z=y+1.

Then

dzdx=dydx,\frac{dz}{dx}=\frac{dy}{dx},dxdz​=dxdy​,

and the DE becomes

dzdx=z(zex2/2−x)=z2ex2/2−xz.\frac{dz}{dx}=z\left(ze^{x^2/2}-x\right)=z^2 e^{x^2/2}-xz.dxdz​=z(zex2/2−x)=z2ex2/2−xz.

So,

dzdx+xz=z2ex2/2.\frac{dz}{dx}+xz=z^2 e^{x^2/2}.dxdz​+xz=z2ex2/2.

This is a Bernoulli-type equation.

  1. Convert using u=1zu=\dfrac{1}{z}u=z1​

Let

u=1z.u=\frac{1}{z}.u=z1​.

Then

dνdx=−1z2dzdx.\frac{d\nu}{dx}=-\frac{1}{z^2}\frac{dz}{dx}.dxdν​=−z21​dxdz​.

From

dzdx=z2ex2/2−xz,\frac{dz}{dx}=z^2 e^{x^2/2}-xz,dxdz​=z2ex2/2−xz,

we get

dνdx=−ex2/2+xν.\frac{d\nu}{dx}=-e^{x^2/2}+x\nu.dxdν​=−ex2/2+xν.

Thus,

dνdx−xν=−ex2/2.\frac{d\nu}{dx}-x\nu=-e^{x^2/2}.dxdν​−xν=−ex2/2.

This is linear in ν\nuν.

  1. Solve the linear equation

The integrating factor is

IF=e∫−x dx=e−x2/2.\mathrm{IF}=e^{\int -x\,dx}=e^{-x^2/2}.IF=e∫−xdx=e−x2/2.

Multiplying the equation by the integrating factor:

e−x2/2dνdx−xe−x2/2ν=−1.e^{-x^2/2}\frac{d\nu}{dx}-xe^{-x^2/2}\nu=-1.e−x2/2dxdν​−xe−x2/2ν=−1.

So,

ddx(νe−x2/2)=−1.\frac{d}{dx}\left(\nu e^{-x^2/2}\right)=-1.dxd​(νe−x2/2)=−1.

Integrating,

νe−x2/2=−x+C.\nu e^{-x^2/2}=-x+C.νe−x2/2=−x+C.

Hence,

ν=ex2/2(C−x).\nu=e^{x^2/2}(C-x).ν=ex2/2(C−x).

Since ν=1z\nu=\dfrac{1}{z}ν=z1​ and z=y+1z=y+1z=y+1,

1y+1=ex2/2(C−x).\frac{1}{y+1}=e^{x^2/2}(C-x).y+11​=ex2/2(C−x).

Therefore,

y+1=1ex2/2(C−x)=e−x2/2C−x.y+1=\frac{1}{e^{x^2/2}(C-x)}=\frac{e^{-x^2/2}}{C-x}.y+1=ex2/2(C−x)1​=C−xe−x2/2​.
  1. Use the condition y(2)=0y(2)=0y(2)=0

At x=2x=2x=2, y=0y=0y=0, so y+1=1y+1=1y+1=1. Thus,

1=e−2C−2.1=\frac{e^{-2}}{C-2}.1=C−2e−2​.

Therefore,

C−2=e−2⇒C=2+e−2.C-2=e^{-2} \quad\Rightarrow\quad C=2+e^{-2}.C−2=e−2⇒C=2+e−2.

So,

y+1=e−x2/22+e−2−x.y+1=\frac{e^{-x^2/2}}{2+e^{-2}-x}.y+1=2+e−2−xe−x2/2​.

At x=1x=1x=1,

y(1)+1=e−1/21+e−2.y(1)+1=\frac{e^{-1/2}}{1+e^{-2}}.y(1)+1=1+e−2e−1/2​.

Multiply numerator and denominator by e2e^2e2:

y(1)+1=e−1/2e2e2+1=e3/2e2+1.y(1)+1=\frac{e^{-1/2}e^2}{e^2+1}=\frac{e^{3/2}}{e^2+1}.y(1)+1=e2+1e−1/2e2​=e2+1e3/2​.
  1. Now compute dydx∣x=1\left.\dfrac{dy}{dx}\right|_{x=1}dxdy​​x=1​

Using the original DE,

dydx=(y+1)((y+1)ex2/2−x).\frac{dy}{dx}=(y+1)\left((y+1)e^{x^2/2}-x\right).dxdy​=(y+1)((y+1)ex2/2−x).

At x=1x=1x=1,

dydx∣x=1=(y(1)+1)((y(1)+1)e1/2−1).\left.\frac{dy}{dx}\right|_{x=1}=(y(1)+1)\left((y(1)+1)e^{1/2}-1\right).dxdy​​x=1​=(y(1)+1)((y(1)+1)e1/2−1).

Now,

y(1)+1=e3/2e2+1.y(1)+1=\frac{e^{3/2}}{e^2+1}.y(1)+1=e2+1e3/2​.

Hence,

(y(1)+1)e1/2=e3/2e1/2e2+1=e2e2+1.(y(1)+1)e^{1/2}=\frac{e^{3/2}e^{1/2}}{e^2+1}=\frac{e^2}{e^2+1}.(y(1)+1)e1/2=e2+1e3/2e1/2​=e2+1e2​.

Therefore,

(y(1)+1)e1/2−1=e2e2+1−1=e2−(e2+1)e2+1=−1e2+1.(y(1)+1)e^{1/2}-1=\frac{e^2}{e^2+1}-1=\frac{e^2-(e^2+1)}{e^2+1}=-\frac{1}{e^2+1}.(y(1)+1)e1/2−1=e2+1e2​−1=e2+1e2−(e2+1)​=−e2+11​.

So,

dydx∣x=1=e3/2e2+1(−1e2+1)=−e3/2(e2+1)2.\left.\frac{dy}{dx}\right|_{x=1}=\frac{e^{3/2}}{e^2+1}\left(-\frac{1}{e^2+1}\right) =-\frac{e^{3/2}}{(e^2+1)^2}.dxdy​​x=1​=e2+1e3/2​(−e2+11​)=−(e2+1)2e3/2​.
  1. Match with the options

This is exactly

−e3/2(e2+1)2,-\frac{e^{3/2}}{(e^2+1)^2},−(e2+1)2e3/2​,

which corresponds to Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

So they agree.

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