What is a general solution to the differential equation y'=sqrtxe^(4y)?

1 Answer
Jul 19, 2016

y = -1/4 ln(4(2/3 x^(3/2) + C))

Explanation:

This is a separable differential equation.

dy/(dx) = x^(1/2) e^(4y)

1/(e^(4y)) * dy/(dx) = x^(1/2)

e^(-4y) dy = x^(1/2) dx

int e^(-4y) dy = int x^(1/2) dx

-1/4 e^(-4y) = 2/3 x^(3/2) + C

-4y = ln(-4(2/3 x^(3/2) + C))

y = -1/4 ln(4(C -2/3 x^(3/2) ))