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

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Answer 1 · 2016-07-19T17:36:54

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

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) ))#