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Differential Equations question

2019 · 8 Apr · Shift 1 · Q38
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  5. /2019 · 8 Apr · Shift 1 · Q38

Differential Equations question

2019 · 8 Apr · Shift 1 · Q38

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1{({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1(x2+1)2dxdy​+2x(x2+1)y=1 such that y(0) = 0. If ay(1)\sqrt ay(1)a​y(1)=π32\pi \over 3232π​ , then the value of 'a' is :
  1. A
    12{1 \over 2}21​
  2. B
    116{1 \over 16}161​
  3. C
    1
  4. D
    14{1 \over 4}41​
View written solutionFree

Correct answer: B

  1. Rewrite the differential equation in linear form

Given

(x2+1)2dydx+2x(x2+1)y=1.(x^2+1)^2\frac{dy}{dx}+2x(x^2+1)y=1.(x2+1)2dxdy​+2x(x2+1)y=1.

Divide throughout by (x2+1)2(x^2+1)^2(x2+1)2:

dydx+2xx2+1y=1(x2+1)2.\frac{dy}{dx}+\frac{2x}{x^2+1}y=\frac{1}{(x^2+1)^2}.dxdy​+x2+12x​y=(x2+1)21​.

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)=2xx2+1,Q(x)=1(x2+1)2.P(x)=\frac{2x}{x^2+1},\qquad Q(x)=\frac{1}{(x^2+1)^2}.P(x)=x2+12x​,Q(x)=(x2+1)21​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x)dx=e∫2xx2+1dx.\text{I.F.}=e^{\int P(x)dx}=e^{\int \frac{2x}{x^2+1}dx}.I.F.=e∫P(x)dx=e∫x2+12x​dx.

Now,

∫2xx2+1dx=ln⁡(x2+1).\int \frac{2x}{x^2+1}dx=\ln(x^2+1).∫x2+12x​dx=ln(x2+1).

So,

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

Multiplying by (x2+1)(x^2+1)(x2+1):

(x2+1)dydx+2xy=1x2+1.(x^2+1)\frac{dy}{dx}+2xy=\frac{1}{x^2+1}.(x2+1)dxdy​+2xy=x2+11​.

The left-hand side is

ddx((x2+1)y).\frac{d}{dx}\big((x^2+1)y\big).dxd​((x2+1)y).

Hence,

ddx((x2+1)y)=1x2+1.\frac{d}{dx}\big((x^2+1)y\big)=\frac{1}{x^2+1}.dxd​((x2+1)y)=x2+11​.
  1. Integrate both sides

Integrating,

(x2+1)y=∫1x2+1 dx=tan⁡−1x+C.(x^2+1)y=\int \frac{1}{x^2+1}\,dx=\tan^{-1}x+C.(x2+1)y=∫x2+11​dx=tan−1x+C.

Thus,

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

At x=0x=0x=0,

0=y(0)=tan⁡−1(0)+C1=C.0=y(0)=\frac{\tan^{-1}(0)+C}{1}=C.0=y(0)=1tan−1(0)+C​=C.

So C=0C=0C=0.

Therefore,

y(x)=tan⁡−1xx2+1.y(x)=\frac{\tan^{-1}x}{x^2+1}.y(x)=x2+1tan−1x​.
  1. Find y(1)y(1)y(1)
y(1)=tan⁡−1(1)12+1=π/42=π8.y(1)=\frac{\tan^{-1}(1)}{1^2+1}=\frac{\pi/4}{2}=\frac{\pi}{8}.y(1)=12+1tan−1(1)​=2π/4​=8π​.
  1. Use the condition a y(1)=π32\sqrt{a}\,y(1)=\dfrac{\pi}{32}a​y(1)=32π​

Substitute y(1)=π8y(1)=\dfrac{\pi}{8}y(1)=8π​:

a⋅π8=π32.\sqrt{a}\cdot \frac{\pi}{8}=\frac{\pi}{32}.a​⋅8π​=32π​.

Cancel π\piπ:

a⋅18=132.\sqrt{a}\cdot \frac{1}{8}=\frac{1}{32}.a​⋅81​=321​.

So,

a=832=14.\sqrt{a}=\frac{8}{32}=\frac{1}{4}.a​=328​=41​.

Squaring both sides,

a=116.a=\frac{1}{16}.a=161​.
  1. Match with the options
a=116a=\frac{1}{16}a=161​

which is Option B.

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