Question

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

• A. \( \frac{1}{4} \left( e^{\pi/2} - 1 \right) \)
• B. \( \frac{1}{2} \left( 1 - e^{\pi/2} \right) \)
• C. \( \frac{1}{4} \left( 1 - e^{\pi/2} \right) \)
• D. \( \frac{1}{2} \left( e^{\pi/2} - 1 \right) \)

Answer 1

The Correct Option is B

The provided differential equation is identified as a first-order linear differential equation and is presented in the form:

\[(1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1}x}.\]

To facilitate solving, the equation is rearranged into its standard form:

\[\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1}x}}{1 + x^2}.\]

The integrating factor (IF) is calculated as:

\[IF = e^{\int \frac{1}{1+x^2} \, dx}= e^{\tan^{-1}x}.\]

Multiplying the differential equation by the integrating factor yields:

\[e^{\tan^{-1}x}\left(\frac{dy}{dx} + \frac{y}{1 + x^2}\right) = \frac{e^{2\tan^{-1}x}}{1 + x^2}.\]

This equation simplifies to the derivative of a product:

\[\frac{d}{dx}\left(ye^{\tan^{-1}x}\right) = \frac{e^{2\tan^{-1}x}}{1 + x^2}.\]

Integration of both sides results in:

\[ye^{\tan^{-1}x} = \int \frac{e^{2\tan^{-1}x}}{1 + x^2} \, dx + C.\]

The integral is evaluated by substituting \(t = \tan^{-1}x\), which implies \(\frac{dt}{dx} = \frac{1}{1+x^2}\) or \(dx = (1 + x^2) \, dt\). The integral then becomes:

\[\int e^{2t} \, dt = \frac{e^{2t}}{2} + C.\]

Substituting back the original variable gives:

\[ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} + C.\]

The initial condition \(y(1) = 0\) at \(x = 1\) is applied:

\[0 = \frac{e^{\pi/4}}{2} + C.\]

Solving for C yields:

\[C = -\frac{e^{\pi/2}}{2}.\]

Substituting the value of C back into the solution provides:

\[ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}.\]

Solving for y gives the explicit solution:

\[y = \frac{e^{\tan^{-1}x}}{2} - \frac{e^{\tan^{-1}x-\pi/2}}{2}.\]

To find \(y(0)\), substitute \(x = 0\):

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

Therefore, the value of \(y(0)\) is:

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