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Differential Equations question

2025 · 24 Jan · Shift 1 · Q34
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  5. /2025 · 24 Jan · Shift 1 · Q34

Differential Equations question

2025 · 24 Jan · Shift 1 · Q34

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)\mathrm{y}=\mathrm{y}(\mathrm{x})y=y(x) be the solution of the differential equation (xy−5x21+x2)dx+(1+x2)dy=0,y(0)=0\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0(xy−5x21+x2​)dx+(1+x2)dy=0,y(0)=0. Then y(3)y(\sqrt{3})y(3​) is equal to
  1. A
    532\frac{5 \sqrt{3}}{2}253​​
  2. B
    152\sqrt{\frac{15}{2}}215​​
  3. C
    143\sqrt{\frac{14}{3}}314​​
  4. D
    222 \sqrt{2}22​
View written solutionFree

Correct answer: A

  1. Write the differential equation in standard form

Given

(xy−5x21+x2) dx+(1+x2) dy=0.(xy-5x^2\sqrt{1+x^2})\,dx+(1+x^2)\,dy=0.(xy−5x21+x2​)dx+(1+x2)dy=0.

So,

(1+x2)dydx+xy−5x21+x2=0.(1+x^2)\frac{dy}{dx}+xy-5x^2\sqrt{1+x^2}=0.(1+x2)dxdy​+xy−5x21+x2​=0.

Hence,

dydx+x1+x2y=5x21+x2.\frac{dy}{dx}+\frac{x}{1+x^2}y=\frac{5x^2}{\sqrt{1+x^2}}.dxdy​+1+x2x​y=1+x2​5x2​.

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)=x1+x2,Q(x)=5x21+x2.P(x)=\frac{x}{1+x^2}, \qquad Q(x)=\frac{5x^2}{\sqrt{1+x^2}}.P(x)=1+x2x​,Q(x)=1+x2​5x2​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x) dx=e∫x1+x2 dx.\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \frac{x}{1+x^2}\,dx}.I.F.=e∫P(x)dx=e∫1+x2x​dx.

Now,

∫x1+x2 dx=12ln⁡(1+x2).\int \frac{x}{1+x^2}\,dx=\frac12\ln(1+x^2).∫1+x2x​dx=21​ln(1+x2).

Therefore,

I.F.=e12ln⁡(1+x2)=1+x2.\text{I.F.}=e^{\frac12\ln(1+x^2)}=\sqrt{1+x^2}.I.F.=e21​ln(1+x2)=1+x2​.
  1. Multiply the equation by the integrating factor

Multiplying throughout by 1+x2\sqrt{1+x^2}1+x2​,

1+x2dydx+x1+x2y=5x2.\sqrt{1+x^2}\frac{dy}{dx}+\frac{x}{\sqrt{1+x^2}}y=5x^2.1+x2​dxdy​+1+x2​x​y=5x2.

The left-hand side is

ddx(y1+x2).\frac{d}{dx}\left(y\sqrt{1+x^2}\right).dxd​(y1+x2​).

So,

ddx(y1+x2)=5x2.\frac{d}{dx}\left(y\sqrt{1+x^2}\right)=5x^2.dxd​(y1+x2​)=5x2.
  1. Integrate both sides

Integrating,

y1+x2=∫5x2 dx=5x33+C.y\sqrt{1+x^2}=\int 5x^2\,dx=\frac{5x^3}{3}+C.y1+x2​=∫5x2dx=35x3​+C.

Thus,

y=5x33+C1+x2.y=\frac{\frac{5x^3}{3}+C}{\sqrt{1+x^2}}.y=1+x2​35x3​+C​.
  1. Use the initial condition y(0)=0y(0)=0y(0)=0

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

0=0+C1  ⟹  C=0.0=\frac{0+C}{1} \implies C=0.0=10+C​⟹C=0.

Therefore,

y=5x331+x2.y=\frac{5x^3}{3\sqrt{1+x^2}}.y=31+x2​5x3​.
  1. Find y(3)y(\sqrt{3})y(3​)

Substitute x=3x=\sqrt{3}x=3​:

y(3)=5(3)331+3.y(\sqrt{3})=\frac{5(\sqrt{3})^3}{3\sqrt{1+3}}.y(3​)=31+3​5(3​)3​.

Since

(3)3=33,4=2,(\sqrt{3})^3=3\sqrt{3}, \qquad \sqrt{4}=2,(3​)3=33​,4​=2,

we get

y(3)=5⋅333⋅2=532.y(\sqrt{3})=\frac{5\cdot 3\sqrt{3}}{3\cdot 2}=\frac{5\sqrt{3}}{2}.y(3​)=3⋅25⋅33​​=253​​.
  1. Check options
y(3)=532\boxed{y(\sqrt{3})=\frac{5\sqrt{3}}{2}}y(3​)=253​​​

So the correct option is A.

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