Using separation of variables, solve the following differential equation with initial conditions dy/dx = e^(2x+3y) and y(0) = 1.

Question

Hint: use a property of exponentials to rewrite the differential equation so it can be separated

Answer 1

dy/dx = e^(2x+3y)
dy/dx = e^(2x)*e^(3y)
e^(-3y) dy = e^(2x) dx
-1/3 e^(-3y) = 1/2 e^(2x) + c
e^(-3y) = -3/2 e^(2x) + c

y(0) = 1, so
e^(-3) = -3/2 + c
c = 3/2 + 1/e^3

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

or, if you insist,

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

Note that the domain is restricted to where

-3/2 e^(2x) + 3/2 + 1/e^3 > 0
x < 1/2 (ln(2/3 + e^3)-3) ≈ 0.016

Answer 2

This is not how you do it. There are examples and Chegg...

Answer 3

Steve is actually right in this problem, he just did it in a different way. Extrapolating based of what he did, you could multiple both sides of the function by -1/3 to get -1/3(e^(-3y))=1/2(e^(2x))-1/2-1/3(e^(-3)) as the answer to the problem