Question

The solution of the differential equation \((1+x) \frac{\mathrm{d} y}{\mathrm{~d} x}-x y=1-x\) is

• A. \(y(1+x)=x+\mathrm{ce}^{x}\), where c is the constant of integration
• B. \(y(1+x)=\mathrm{ce}^{x}\), where c is the constant of integration
• C. \(y(1-x)=x-\mathrm{ce}^{x}\), where c is the constant of integration
• D. \(y(1+x)=x\), where c is the constant of integration

Hint:

Write it as a linear equation, find the integrating factor, and integrate once.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The equation is linear: \(\dfrac{dy}{dx}-\dfrac{x}{1+x}y=\dfrac{1-x}{1+x}\).

Step 2: Integrating factor
\(\int-\dfrac{x}{1+x}dx=-x+\log(1+x)\), so \(\text{I.F.}=e^{-x}(1+x)\).

Step 3: Integrate
\(y\,e^{-x}(1+x)=\int\dfrac{1-x}{1+x}\cdot e^{-x}(1+x)\,dx=\int(1-x)e^{-x}dx\).
Since \(\dfrac{d}{dx}(xe^{-x})=(1-x)e^{-x}\), the integral is \(xe^{-x}+c\).

Step 4: Result
\(y(1+x)=x+ce^{x}\).

Final Answer:
The solution is y(1+x) = x + c e^x. \[ \boxed{\text{(A) }y(1+x)=x+\mathrm{c}\mathrm{e}^{x}} \]