Question

The solution of the differential equation $(1+x)\frac{dy}{dx}-xy=1-x$ is}

• A. $y(1+x)=x+ce^{x}$
• B. $y(1+x)=ce^{x}$
• C. $y(1-x)=x-ce^{x}$
• D. $y(1+x)=x$

Hint:

In Linear DE, the I.F. usually simplifies the RHS significantly.

Answer 1

The Correct Option is A

Step 1: DE Form
Divide by $(1+x)$: $\frac{dy}{dx} - \frac{x}{1+x}y = \frac{1-x}{1+x}$.

Step 2: Integrating Factor
$P = - \frac{x}{1+x} = - (1 - \frac{1}{1+x})$.
$I.F. = e^{\int (-1 + \frac{1}{1+x}) dx} = e^{-x + \log(1+x)} = (1+x)e^{-x}$.

Step 3: Calculation
$y \cdot (1+x)e^{-x} = \int \frac{1-x}{1+x} (1+x)e^{-x} dx = \int (1-x)e^{-x} dx$.
Using parts: $\int (1-x)e^{-x} dx = (x-1)e^{-x} - \int e^{-x} dx = (x-1)e^{-x} + e^{-x} + C = xe^{-x} + C$.
$y(1+x)e^{-x} = xe^{-x} + C$.
$y(1+x) = x + Ce^x$.

Step 4: Conclusion
Hence, $y(1+x) = x + ce^x$. Final Answer: (A)