Question

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

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

Hint:

For separable differential equations, isolate variables and integrate both sides, ensuring to adjust the constant of integration appropriately.

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{dy}{dx} = 2x \cdot e^{x^2 + y} \] This is a first-order differential equation. To solve it, we can use the method of separation of variables. Rewrite the equation by separating the variables \( y \) and \( x \): \[ \frac{dy}{dx} = 2x \cdot e^{x^2} \cdot e^y \] Rearrange to isolate terms involving \( y \) and \( x \): \[ \frac{dy}{e^y} = 2x \cdot e^{x^2} \, dx \] Now, integrate both sides: \[ \int e^{-y} \, dy = \int 2x \cdot e^{x^2} \, dx \] The left-hand side is: \[ \int e^{-y} \, dy = -e^{-y} \] For the right-hand side, use substitution. Let \( u = x^2 \), so \( du = 2x \, dx \), and the integral becomes: \[ \int 2x \cdot e^{x^2} \, dx = \int e^u \, du = e^u + C_1 = e^{x^2} + C_1 \] Thus, we have: \[ -e^{-y} = e^{x^2} + C_1 \] Multiply through by \(-1\) to simplify: \[ e^{-y} = -e^{x^2} - C_1 \] Since \( C_1 \) is an arbitrary constant, let \( C = -C_1 \) (adjusting the constant): \[ e^{-y} = -e^{x^2} + C \] Rearrange to combine terms: \[ e^{-y} + e^{x^2} = C \] This is the general solution. To match the form in the options, note that: \[ e^{x^2} + e^{-y} = C \] This matches option (B). \[ \boxed{e^{x^2} + e^{-y} = C} \]