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Differential Equations question

2023 · 13 Apr · Shift 2 · Q41
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  5. /2023 · 13 Apr · Shift 2 · Q41

Differential Equations question

2023 · 13 Apr · Shift 2 · Q41

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
If y=y(x)y=y(x)y=y(x) is the solution of the differential equation dydx+4x(x2−1)y=x+2(x2−1)52,x>1\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x \gt 1dxdy​+(x2−1)4x​y=(x2−1)25​x+2​,x>1 such that y(2)=29log⁡e(2+3) and y(2)=αlog⁡e(α+β)+β−γ,α,β,γ∈N, then αβγ is equal to y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }y(2)=92​loge​(2+3​) and y(2​)=αloge​(α​+β)+β−γ​,α,β,γ∈N, then αβγ is equal to  :
Numerical answer
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Correct answer: 6

  1. Given differential equation
dydx+4xx2−1y=x+2(x2−1)5/2,x>1\frac{dy}{dx}+\frac{4x}{x^2-1}y=\frac{x+2}{(x^2-1)^{5/2}}, \qquad x>1dxdy​+x2−14x​y=(x2−1)5/2x+2​,x>1

This is a linear differential equation of the form

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

with

P(x)=4xx2−1,Q(x)=x+2(x2−1)5/2.P(x)=\frac{4x}{x^2-1}, \qquad Q(x)=\frac{x+2}{(x^2-1)^{5/2}}.P(x)=x2−14x​,Q(x)=(x2−1)5/2x+2​.
  1. Find the integrating factor
I.F.=e∫P(x) dx=e∫4xx2−1 dx\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \frac{4x}{x^2-1}\,dx}I.F.=e∫P(x)dx=e∫x2−14x​dx

Let u=x2−1u=x^2-1u=x2−1, then du=2x dxdu=2x\,dxdu=2xdx. Hence,

∫4xx2−1 dx=2∫2xx2−1 dx=2ln⁡(x2−1)\int \frac{4x}{x^2-1}\,dx=2\int \frac{2x}{x^2-1}\,dx=2\ln(x^2-1)∫x2−14x​dx=2∫x2−12x​dx=2ln(x2−1)

since x>1x>1x>1, x2−1>0x^2-1>0x2−1>0.

Thus,

I.F.=e2ln⁡(x2−1)=(x2−1)2.\text{I.F.}=e^{2\ln(x^2-1)}=(x^2-1)^2.I.F.=e2ln(x2−1)=(x2−1)2.
  1. Multiply the equation by the integrating factor
(x2−1)2dydx+4xx2−1(x2−1)2y=(x2−1)2⋅x+2(x2−1)5/2(x^2-1)^2\frac{dy}{dx}+\frac{4x}{x^2-1}(x^2-1)^2y=(x^2-1)^2\cdot \frac{x+2}{(x^2-1)^{5/2}}(x2−1)2dxdy​+x2−14x​(x2−1)2y=(x2−1)2⋅(x2−1)5/2x+2​

So,

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

Hence,

(x2−1)2y=∫x+2x2−1 dx+C.(x^2-1)^2y=\int \frac{x+2}{\sqrt{x^2-1}}\,dx+C.(x2−1)2y=∫x2−1​x+2​dx+C.
  1. Evaluate the integral

Split it as:

∫x+2x2−1dx=∫xx2−1dx+2∫1x2−1dx.\int \frac{x+2}{\sqrt{x^2-1}}dx=\int \frac{x}{\sqrt{x^2-1}}dx+2\int \frac{1}{\sqrt{x^2-1}}dx.∫x2−1​x+2​dx=∫x2−1​x​dx+2∫x2−1​1​dx.

Now,

∫xx2−1dx=x2−1\int \frac{x}{\sqrt{x^2-1}}dx=\sqrt{x^2-1}∫x2−1​x​dx=x2−1​

and for x>1x>1x>1,

∫1x2−1dx=ln⁡(x+x2−1).\int \frac{1}{\sqrt{x^2-1}}dx=\ln\left(x+\sqrt{x^2-1}\right).∫x2−1​1​dx=ln(x+x2−1​).

Therefore,

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

So,

y=x2−1+2ln⁡(x+x2−1)+C(x2−1)2.y=\frac{\sqrt{x^2-1}+2\ln\left(x+\sqrt{x^2-1}\right)+C}{(x^2-1)^2}.y=(x2−1)2x2−1​+2ln(x+x2−1​)+C​.
  1. Use the condition y(2)=29ln⁡(2+3)y(2)=\dfrac{2}{9}\ln(2+\sqrt{3})y(2)=92​ln(2+3​)

At x=2x=2x=2,

x2−1=3,x2−1=3.x^2-1=3, \qquad \sqrt{x^2-1}=\sqrt{3}.x2−1=3,x2−1​=3​.

Hence,

y(2)=3+2ln⁡(2+3)+C9.y(2)=\frac{\sqrt{3}+2\ln(2+\sqrt{3})+C}{9}.y(2)=93​+2ln(2+3​)+C​.

Given,

3+2ln⁡(2+3)+C9=29ln⁡(2+3).\frac{\sqrt{3}+2\ln(2+\sqrt{3})+C}{9}=\frac{2}{9}\ln(2+\sqrt{3}).93​+2ln(2+3​)+C​=92​ln(2+3​).

Multiply by 999:

3+2ln⁡(2+3)+C=2ln⁡(2+3).\sqrt{3}+2\ln(2+\sqrt{3})+C=2\ln(2+\sqrt{3}).3​+2ln(2+3​)+C=2ln(2+3​).

Thus,

C=−3.C=-\sqrt{3}.C=−3​.

So the solution is

y=x2−1+2ln⁡(x+x2−1)−3(x2−1)2.y=\frac{\sqrt{x^2-1}+2\ln\left(x+\sqrt{x^2-1}\right)-\sqrt{3}}{(x^2-1)^2}.y=(x2−1)2x2−1​+2ln(x+x2−1​)−3​​.
  1. Find y(2)y(\sqrt{2})y(2​)

At x=2x=\sqrt{2}x=2​,

x2−1=1,x2−1=1,(x2−1)2=1.x^2-1=1, \qquad \sqrt{x^2-1}=1, \qquad (x^2-1)^2=1.x2−1=1,x2−1​=1,(x2−1)2=1.

Therefore,

y(2)=1+2ln⁡(2+1)−3.y(\sqrt{2})=1+2\ln(\sqrt{2}+1)-\sqrt{3}.y(2​)=1+2ln(2​+1)−3​.

So,

y(2)=2ln⁡(1+2)+1−3.y(\sqrt{2})=2\ln(1+\sqrt{2})+1-\sqrt{3}.y(2​)=2ln(1+2​)+1−3​.

This is given in the form

αln⁡(α+β)+β−γ.\alpha \ln(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}.αln(α​+β)+β−γ​.

Comparing,

α=2,β=1,γ=3.\alpha=2, \quad \beta=1, \quad \gamma=3.α=2,β=1,γ=3.

Thus,

αβγ=2⋅1⋅3=6.\alpha\beta\gamma=2\cdot 1\cdot 3=6.αβγ=2⋅1⋅3=6.
  1. Comparison with stored answer

Derived answer = 666.

Stored correct answer = 666.

They agree.

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