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 given differential equation is:

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

Integrating factor (I.F.):

\[ \text{I.F.} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x}. \]

Multiply through by I.F.:

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

Simplify:

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

Substitute:

\[ \tan^{-1}x = z, \quad \frac{1}{1+x^2} dx = dz. \]

Thus:

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

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

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

Finally:

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

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

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