Using separation of variables, solve the following differential equation with initial conditions dy/dx = e^(2x+3y) and y(0) = 1.
Hint: use a property of exponentials to rewrite the differential equation so it can be separated
3 answers

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

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

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))1/21/3(e^(3)) as the answer to the problem
