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Differential Equations question

2024 · 30 Jan · Shift 1 · Q52
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  5. /2024 · 30 Jan · Shift 1 · Q52

Differential Equations question

2024 · 30 Jan · Shift 1 · Q52

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1−x2)dy=[xy+(x3+2)3(1−x2)]dx,−1<x<1,y(0)=0\left(1-x^2\right) \mathrm{d} y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] \mathrm{d} x, -1\lt x\lt 1, y(0)=0(1−x2)dy=[xy+(x3+2)3(1−x2)​]dx,−1<x<1,y(0)=0. If y(12)=mn,my\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}y(21​)=nm​,m and n\mathrm{n}n are co-prime numbers, then m+n\mathrm{m}+\mathrm{n}m+n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 97

We need to solve the differential equation

(1−x2) dy=[xy+(x3+2)3(1−x2)]dx,−1<x<1,y(0)=0.(1-x^2)\,dy=\left[xy+(x^3+2)\sqrt{3(1-x^2)}\right]dx, \qquad -1<x<1, \qquad y(0)=0.(1−x2)dy=[xy+(x3+2)3(1−x2)​]dx,−1<x<1,y(0)=0.

We want y(12)y\left(\frac12\right)y(21​).

1. Rewrite in standard linear form

Divide by dxdxdx:

(1−x2)dydx=xy+(x3+2)3(1−x2).(1-x^2)\frac{dy}{dx}=xy+(x^3+2)\sqrt{3(1-x^2)}.(1−x2)dxdy​=xy+(x3+2)3(1−x2)​.

So

dydx−x1−x2y=(x3+2)3(1−x2)1−x2=3 x3+21−x2.\frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{(x^3+2)\sqrt{3(1-x^2)}}{1-x^2} =\sqrt{3}\,\frac{x^3+2}{\sqrt{1-x^2}}.dxdy​−1−x2x​y=1−x2(x3+2)3(1−x2)​​=3​1−x2​x3+2​.

Thus the linear differential equation is

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)=3 x3+21−x2.P(x)=-\frac{x}{1-x^2}, \qquad Q(x)=\sqrt{3}\,\frac{x^3+2}{\sqrt{1-x^2}}.P(x)=−1−x2x​,Q(x)=3​1−x2​x3+2​.

2. Find the integrating factor

The integrating factor is

IF=e∫P(x)dx=e∫−x1−x2dx.IF=e^{\int P(x)dx}=e^{\int -\frac{x}{1-x^2}dx}.IF=e∫P(x)dx=e∫−1−x2x​dx.

Now,

∫−x1−x2dx=12ln⁡(1−x2),\int -\frac{x}{1-x^2}dx =\frac12\ln(1-x^2),∫−1−x2x​dx=21​ln(1−x2),

because

ddx(1−x2)=−2x.\frac{d}{dx}(1-x^2)=-2x.dxd​(1−x2)=−2x.

Hence

IF=e12ln⁡(1−x2)=1−x2.IF=e^{\frac12\ln(1-x^2)}=\sqrt{1-x^2}.IF=e21​ln(1−x2)=1−x2​.

3. Multiply the equation by the integrating factor

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

1−x2 dydx−x1−x2y=3(x3+2).\sqrt{1-x^2}\,\frac{dy}{dx}-\frac{x}{\sqrt{1-x^2}}y =\sqrt{3}(x^3+2).1−x2​dxdy​−1−x2​x​y=3​(x3+2).

The left-hand side is

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

Therefore,

ddx(y1−x2)=3(x3+2).\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=\sqrt{3}(x^3+2).dxd​(y1−x2​)=3​(x3+2).

4. Integrate

Integrating,

y1−x2=3∫(x3+2) dx+C.y\sqrt{1-x^2}=\sqrt{3}\int (x^3+2)\,dx + C.y1−x2​=3​∫(x3+2)dx+C.

So

y1−x2=3(x44+2x)+C.y\sqrt{1-x^2}=\sqrt{3}\left(\frac{x^4}{4}+2x\right)+C.y1−x2​=3​(4x4​+2x)+C.

Use the condition y(0)=0y(0)=0y(0)=0:

0⋅1−0=3(0)+C  ⟹  C=0.0\cdot \sqrt{1-0}=\sqrt{3}(0)+C \implies C=0.0⋅1−0​=3​(0)+C⟹C=0.

Thus

y=3(x44+2x)1−x2.y=\frac{\sqrt{3}\left(\frac{x^4}{4}+2x\right)}{\sqrt{1-x^2}}.y=1−x2​3​(4x4​+2x)​.

5. Evaluate at x=12x=\frac12x=21​

First,

x44+2x=(1/2)44+2⋅12=1/164+1=164+1=6564.\frac{x^4}{4}+2x =\frac{(1/2)^4}{4}+2\cdot\frac12 =\frac{1/16}{4}+1 =\frac{1}{64}+1 =\frac{65}{64}.4x4​+2x=4(1/2)4​+2⋅21​=41/16​+1=641​+1=6465​.

Also,

1−x2=1−14=34=32.\sqrt{1-x^2}=\sqrt{1-\frac14}=\sqrt{\frac34}=\frac{\sqrt{3}}{2}.1−x2​=1−41​​=43​​=23​​.

Hence

y(12)=3⋅656432=6564⋅2=6532.y\left(\frac12\right)=\frac{\sqrt{3}\cdot \frac{65}{64}}{\frac{\sqrt{3}}{2}} =\frac{65}{64}\cdot 2 =\frac{65}{32}.y(21​)=23​​3​⋅6465​​=6465​⋅2=3265​.

So

m=65,n=32.m=65,\qquad n=32.m=65,n=32.

Since 656565 and 323232 are co-prime,

m+n=65+32=97.m+n=65+32=97.m+n=65+32=97.

6. Comparison with stored answer

Derived answer: 979797.

Stored correct answer: 979797.

They agree.

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