Let y=y(x) be the solution of the differential equation [(1+x^2) (d y)/(d x)+y=e^ ^-1 x], …

Question

Let $y=y(x)$ be the solution of the differential equation $$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$$, $y(1)=0$. Then $y(0)$ is

Accepted Answer

To determine $ y(0) $, we start by solving the differential equation given:

$ (1 + x^2) \frac{dy}{dx} + y = e^{	an^{-1} x} $

First, we rewrite it in the standard form for a linear differential equation:

$ \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{	an^{-1} x}}{1 + x^2} $

Next, we find the integrating factor (I.F.):

$ 	ext{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{	an^{-1} x} $

Multiply through by the integrating factor:

$ y \cdot e^{	an^{-1} x} = \int e^{	an^{-1} x} \cdot \frac{e^{	an^{-1} x}}{1 + x^2} dx $

This simplifies to:

$ y \cdot e^{	an^{-1} x} = \int \frac{e^{2 	an^{-1} x}}{1 + x^2} dx $

Make the substitution $ 	an^{-1} x = t $, then $ \frac{1}{1 + x^2} dx = dt $:

$ \int e^{2t} dt = \frac{e^{2t}}{2} + \frac{C}{2} $

Rewrite in terms of $ x $:

$ y \cdot e^{	an^{-1} x} = \frac{e^{2 	an^{-1} x}}{2} + \frac{C}{2} $

Use the initial condition $ y(1) = 0 $:

$ 0 = \frac{e^{2 \cdot 	an^{-1} 1}}{2} + \frac{C}{2} $

Since $ 	an^{-1} 1 = \frac{\pi}{4} $:

$ 0 = \frac{e^{\pi/2}}{2} + \frac{C}{2} \implies C = -e^{\pi/2} $

Thus, the solution is:

$ y \cdot e^{	an^{-1} x} = \frac{e^{2 	an^{-1} x} - e^{\pi/2}}{2} $

Evaluating $ y(0) $:

$ y(0) = y \cdot 1 = \frac{1 - e^{\pi/2}}{2} $

Therefore,

$ y(0) = \frac{1}{2} (1 - e^{\pi/2}) $

Hence, the correct answer is:

Option B

$ \frac{1}{2}\left(1-e^{\pi / 2}ight) $

Solutions · 1