Let y=y(x) be the solution of the differential equation (1+x^2)(dy)/(dx)+y=e^ ^-1x, y(1)=0…

Question

Let $y=y(x)$ be the solution of the differential equation $(1+{x}^{2})\frac{dy}{dx}+y={e}^{{	an }^{-1}x}$, $y(1)=0$. 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.):

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

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

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

Solutions · 1