Question

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

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

Hint:

When dealing with mixed exponential terms like \( e^{x-y} \), factoring out the negative exponent component is almost always the fastest way to separate variables cleanly.

Answer 1

The Correct Option is A

Concept: A separable differential equation can be solved by grouping all terms containing the dependent variable \( y \) on one side of the equation and all terms containing the independent variable \( x \) on the other side, allowing you to integrate each side directly: \[ \int h(y)\,dy = \int g(x)\,dx \]

Step 1: Separate the mixed terms using exponent rules. Using the laws of exponents, rewrite the exponential term on the right side: \[ e^{x-y} = e^x \cdot e^{-y} \] Substitute this back into the differential equation: \[ \frac{dy}{dx} = e^x \cdot e^{-y} + x^2 \cdot e^{-y} \] Factor out the common term \( e^{-y} \) from the right side: \[ \frac{dy}{dx} = e^{-y} \left( e^x + x^2 \right) \]

Step 2: Move variables to opposite sides of the equation. Multiply both sides by \( dx \) and divide by \( e^{-y} \) to group the variables: \[ \frac{1}{e^{-y}}\,dy = \left( e^x + x^2 \right)dx \quad \Rightarrow \quad e^y\,dy = \left( e^x + x^2 \right)dx \]

Step 3: Integrate both sides to find the general solution. Set up and evaluate the integrals on both sides: \[ \int e^y\,dy = \int \left( e^x + x^2 \right)dx \] Integrating using standard rules: \[ e^y = e^x + \frac{x^3}{3} + C \]