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Mathematics Question on Differential equations

Let y=y(x)y = y(x) be the solution of the differential equation (1+x2)dydx+y=etan⁡−1x, y(1)=0.\left( 1 + x^2 \right) \frac{dy}{dx} + y = e^{\tan^{-1}x}, \, y(1) = 0.Then y(0)y(0) is:

A

14(eπ/2−1)\frac{1}{4} \left( e^{\pi/2} - 1 \right)

B

12(1−eπ/2)\frac{1}{2} \left( 1 - e^{\pi/2} \right)

C

14(1−eπ/2)\frac{1}{4} \left( 1 - e^{\pi/2} \right)

D

12(eπ/2−1)\frac{1}{2} \left( e^{\pi/2} - 1 \right)

Answer

12(1−eπ/2)\frac{1}{2} \left( 1 - e^{\pi/2} \right)

Explanation

Solution

The given differential equation is:

dydx+y1+x2=etan⁡−1x1+x2.\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\tan^{-1}x}}{1+x^2}.

Integrating factor (I.F.):

I.F.=e∫11+x2dx=etan⁡−1x.\text{I.F.} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x}.

Multiply through by I.F.:

y⋅etan⁡−1x=∫etan⁡−1x⋅etan⁡−1x1+x2dx+C.y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{e^{\tan^{-1}x}}{1+x^2} dx + C.

Simplify:

y⋅etan⁡−1x=∫e2tan⁡−1x⋅11+x2dx+C.y \cdot e^{\tan^{-1}x} = \int e^{2\tan^{-1}x} \cdot \frac{1}{1+x^2} dx + C.

Substitute:

tan⁡−1x=z,11+x2dx=dz.\tan^{-1}x = z, \quad \frac{1}{1+x^2} dx = dz.

Thus:

y⋅etan⁡−1x=e2z2+C.y \cdot e^{\tan^{-1}x} = \frac{e^{2z}}{2} + C.

Apply initial condition: y(1)=0y(1) = 0, tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}:

0=eπ/22+C  ⟹  C=−eπ/22.0 = \frac{e^{\pi/2}}{2} + C \implies C = -\frac{e^{\pi/2}}{2}.

Finally:

y⋅etan⁡−1x=e2tan⁡−1x2−eπ/22.y \cdot e^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}.

At x=0x = 0, tan⁡−1(0)=0\tan^{-1}(0) = 0:

y(0)=12(1−e−π/2).y(0) = \frac{1}{2} \left( 1 - e^{-\pi/2} \right).