Question

The solution of the differential equation \(\frac{dy}{dx} = e^{x-y}+x^2e^{-y}\) is...

• A. \(y = e^x+c\)
• B. \(e^y = e^x+x^3+c\)
• C. \(e^y = e^x+\frac{x^3}{3}+c\)
• D. \(e^y = e^x+2x+c\)

Hint:

Take e^(-y) as a common factor, then multiply by e^y and integrate.

Answer 1

The Correct Option is C

Step 1: Factor the right side:
Write \(e^{x-y} = e^{x}e^{-y}\). Then
\[ \frac{dy}{dx} = e^{-y}\left(e^x + x^2\right) \]
Both terms share the factor \(e^{-y}\), so the variables can be separated.

Step 2: Separate the variables:
Multiply both sides by \(e^{y}\) and by \(dx\):
\[ e^{y}\,dy = \left(e^x + x^2\right)dx \]

Step 3: Integrate both sides:
The left side integrates to \(e^{y}\). On the right, \(\int e^x\,dx = e^x\) and \(\int x^2\,dx = \frac{x^3}{3}\). So
\[ e^{y} = e^{x} + \frac{x^3}{3} + c \]

Step 4: Why the other options are wrong:
Check each option by differentiating: from \(e^y = f(x)\) we get \(e^y\frac{dy}{dx} = f'(x)\), and this must equal \(e^x + x^2\). Option (B) gives \(e^x + 3x^2\), because \(\frac{d}{dx}x^3 = 3x^2\). Option (D) gives \(e^x + 2\). Option (A), \(y = e^x + c\), gives \(\frac{dy}{dx} = e^x\), which does not satisfy the given equation since \(e^{x-y} + x^2e^{-y} \neq e^x\) in general.

Final Answer:
The solution is \(e^y = e^x + \frac{x^3}{3} + c\), which is option (C). \[ \boxed{e^y = e^x + \frac{x^3}{3} + c} \]