Question

The general solution of the differential equation \[ \frac{dy}{dx}=e^{x-y}+x^2e^{-y} \] is

• A. \(e^{-y}=e^x+\frac{x^3}{3}+c\)
• B. \(e^y=e^x+\frac{x^3}{3}+c\)
• C. \(e^y=e^x+x^3+c\)
• D. \(e^y=e^x+c\)

Hint:

If \(e^{-y}\) appears in a differential equation, multiply by \(e^y\) to form \(\frac{d}{dx}(e^y)\).

Answer 1

The Correct Option is B

Concept: When the equation contains \(e^{-y}\), multiply the equation by \(e^y\) because: \[ e^y\frac{dy}{dx}=\frac{d}{dx}(e^y) \]

Step 1: Given: \[ \frac{dy}{dx}=e^{x-y}+x^2e^{-y} \]

Step 2: Rewrite the right side. \[ e^{x-y}=e^x e^{-y} \] So: \[ \frac{dy}{dx}=e^x e^{-y}+x^2e^{-y} \] \[ \frac{dy}{dx}=e^{-y}(e^x+x^2) \]

Step 3: Multiply both sides by \(e^y\). \[ e^y\frac{dy}{dx}=e^x+x^2 \]

Step 4: Observe that: \[ e^y\frac{dy}{dx}=\frac{d}{dx}(e^y) \] Therefore: \[ \frac{d}{dx}(e^y)=e^x+x^2 \]

Step 5: Integrate both sides with respect to \(x\). \[ \int d(e^y)=\int (e^x+x^2)\,dx \] \[ e^y=e^x+\frac{x^3}{3}+c \] Therefore, \[ \boxed{e^y=e^x+\frac{x^3}{3}+c} \]