solve the equation. dy/dx = 8x^3y - 8xy choose 1 answer: a y = ce^(2x^4 - 4x^2) b y = e^(2x^4 - 4x^2) + c c…

Explanation:

Step1: Separate variables

We start with $\frac{dy}{dx}=8x^{3}y - 8xy$. Factor out $8xy$ on the right - hand side to get $\frac{dy}{dx}=8xy(x^{2}-1)$. Then separate variables: $\frac{dy}{y}=8x(x^{2}-1)dx$.

Step2: Integrate both sides

Integrate $\int\frac{dy}{y}=\int8x(x^{2}-1)dx$. Let $u = x^{2}$, then $du=2xdx$. So $\int8x(x^{2}-1)dx = 4\int(u - 1)du$. Integrating, we have $\int\frac{dy}{y}=4\int(u - 1)du$. $\ln|y|=4(\frac{u^{2}}{2}-u)+C$. Substituting back $u = x^{2}$, we get $\ln|y|=4(\frac{x^{4}}{2}-x^{2})+C$. $\ln|y|=2x^{4}-4x^{2}+C$.

Step3: Solve for y

Exponentiate both sides: $y = e^{2x^{4}-4x^{2}+C}=e^{C}e^{2x^{4}-4x^{2}}$. Let $e^{C}=C_{1}$, then $y = C_{1}e^{2x^{4}-4x^{2}}$.

Answer:

A. $y = Ce^{2x^{4}-4x^{2}}$