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Differential Equations question

2024 · 4 Apr · Shift 2 · Q41
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  5. /2024 · 4 Apr · Shift 2 · Q41

Differential Equations question

2024 · 4 Apr · Shift 2 · Q41

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy−2)dx=0(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0(x2+4)2dy+(2x3y+8xy−2)dx=0. If y(0)=0y(0)=0y(0)=0, then y(2)y(2)y(2) is equal to
  1. A
    2π2 \pi2π
  2. B
    π8\frac{\pi}{8}8π​
  3. C
    π16\frac{\pi}{16}16π​
  4. D
    π32\frac{\pi}{32}32π​
View written solutionFree

Correct answer: D

  1. Rewrite the differential equation

Given

(x2+4)2 dy+(2x3y+8xy−2) dx=0.(x^2+4)^2\,dy+(2x^3y+8xy-2)\,dx=0.(x2+4)2dy+(2x3y+8xy−2)dx=0.

Divide by dxdxdx:

(x2+4)2dydx+2x3y+8xy−2=0.(x^2+4)^2\frac{dy}{dx}+2x^3y+8xy-2=0.(x2+4)2dxdy​+2x3y+8xy−2=0.

So,

(x2+4)2dydx+(2x3+8x)y=2.(x^2+4)^2\frac{dy}{dx}+(2x^3+8x)y=2.(x2+4)2dxdy​+(2x3+8x)y=2.

Thus,

dydx+2x3+8x(x2+4)2y=2(x2+4)2.\frac{dy}{dx}+\frac{2x^3+8x}{(x^2+4)^2}y=\frac{2}{(x^2+4)^2}.dxdy​+(x2+4)22x3+8x​y=(x2+4)22​.

This is a linear differential equation:

dydx+P(x)y=Q(x),\frac{dy}{dx}+P(x)y=Q(x),dxdy​+P(x)y=Q(x),

with

P(x)=2x3+8x(x2+4)2=2x(x2+4)(x2+4)2=2xx2+4.P(x)=\frac{2x^3+8x}{(x^2+4)^2}=\frac{2x(x^2+4)}{(x^2+4)^2}=\frac{2x}{x^2+4}.P(x)=(x2+4)22x3+8x​=(x2+4)22x(x2+4)​=x2+42x​.

So the equation becomes

dydx+2xx2+4y=2(x2+4)2.\frac{dy}{dx}+\frac{2x}{x^2+4}y=\frac{2}{(x^2+4)^2}.dxdy​+x2+42x​y=(x2+4)22​.
  1. Find the integrating factor

The integrating factor is

IF=e∫2xx2+4 dx.IF=e^{\int \frac{2x}{x^2+4}\,dx}.IF=e∫x2+42x​dx.

Let u=x2+4u=x^2+4u=x2+4, then du=2x dxdu=2x\,dxdu=2xdx. Hence,

∫2xx2+4 dx=∫duu=ln⁡(x2+4).\int \frac{2x}{x^2+4}\,dx=\int \frac{du}{u}=\ln(x^2+4).∫x2+42x​dx=∫udu​=ln(x2+4).

Therefore,

IF=eln⁡(x2+4)=x2+4.IF=e^{\ln(x^2+4)}=x^2+4.IF=eln(x2+4)=x2+4.
  1. Multiply the equation by the integrating factor

Multiplying throughout by (x2+4)(x^2+4)(x2+4):

(x2+4)dydx+2xy=2x2+4.(x^2+4)\frac{dy}{dx}+2xy=\frac{2}{x^2+4}.(x2+4)dxdy​+2xy=x2+42​.

The left-hand side is

ddx((x2+4)y).\frac{d}{dx}\big((x^2+4)y\big).dxd​((x2+4)y).

So,

ddx((x2+4)y)=2x2+4.\frac{d}{dx}\big((x^2+4)y\big)=\frac{2}{x^2+4}.dxd​((x2+4)y)=x2+42​.
  1. Integrate both sides

Integrating,

(x2+4)y=∫2x2+4 dx+C.(x^2+4)y=\int \frac{2}{x^2+4}\,dx + C.(x2+4)y=∫x2+42​dx+C.

Use the standard formula

∫dxx2+a2=1atan⁡−1(xa).\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right).∫x2+a2dx​=a1​tan−1(ax​).

Here a=2a=2a=2, so

∫2x2+4 dx=2⋅12tan⁡−1(x2)=tan⁡−1(x2).\int \frac{2}{x^2+4}\,dx=2\cdot \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)=\tan^{-1}\left(\frac{x}{2}\right).∫x2+42​dx=2⋅21​tan−1(2x​)=tan−1(2x​).

Thus,

(x2+4)y=tan⁡−1(x2)+C.(x^2+4)y=\tan^{-1}\left(\frac{x}{2}\right)+C.(x2+4)y=tan−1(2x​)+C.

Hence,

y=tan⁡−1(x/2)+Cx2+4.y=\frac{\tan^{-1}(x/2)+C}{x^2+4}.y=x2+4tan−1(x/2)+C​.
  1. Use the initial condition y(0)=0y(0)=0y(0)=0

Substitute x=0x=0x=0, y=0y=0y=0:

0=tan⁡−1(0)+C4=C4.0=\frac{\tan^{-1}(0)+C}{4}=\frac{C}{4}.0=4tan−1(0)+C​=4C​.

So,

C=0.C=0.C=0.

Therefore,

y(x)=tan⁡−1(x/2)x2+4.y(x)=\frac{\tan^{-1}(x/2)}{x^2+4}.y(x)=x2+4tan−1(x/2)​.
  1. Find y(2)y(2)y(2)

Substitute x=2x=2x=2:

y(2)=\frac{\tan^{-1}(1)}{2^2+4}= rac{\pi/4}{8}=\frac{\pi}{32}.
  1. Compare with the options
y(2)=π32.y(2)=\frac{\pi}{32}.y(2)=32π​.

So the correct option is:

D: π32\frac{\pi}{32}32π​

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