Question

The solution for the differential equation $\log \left( \frac{dy}{dx} \right) = 3x + 4y$ is:

• A. $3e^{3y} + 4e^{-3x} + C = 0$
• B. $e^{3x} + 4y + C = 0$
• C. $3e^{-y} + 4e^{x} + 12C = 0$
• D. $3e^{-y} + 4e^{3x} + 12C = 0$

Hint:

When solving separable differential equations, always isolate terms involving $y$ on one side and terms involving $x$ on the other.

Answer 1

The Correct Option is C

The given differential equation is: \[ \log \left( \frac{dy}{dx} \right) = 3x + 4y. \] Taking the exponential of both sides: \[ \frac{dy}{dx} = e^{3x + 4y}. \] Now, separate the variables: \[ \frac{dy}{e^{4y}} = e^{3x} dx. \] Integrating both sides: \[ \int \frac{dy}{e^{4y}} = \int e^{3x} dx. \] The integral of $\frac{dy}{e^{4y}}$ is $-\frac{1}{4e^{4y}}$, and the integral of $e^{3x}$ is $\frac{e^{3x}}{3}$. So we get: \[ -\frac{1}{4e^{4y}} = \frac{e^{3x}}{3} + C. \] Simplifying: \[ 3e^{-4y} + 4e^{3x} + 12C = 0. \] Thus, the solution is $3e^{-y} + 4e^{x} + 12C = 0$.