# Thread: Particular Solution - Calculus based question

1. ## Particular Solution - Calculus based question

So I have a fairly simple looking question I just forgot how to complete it.

2ydx = 3xdy when y(-2) = 1

dx/(3x) = dy(2y)

ln(3x) = ln(2y) + C

This is where I get stuck. I need to solve for y but I've just forgotten how to do this and don't know how to describe this type of problem.
Thanks for any help!

2. Originally Posted by coldize
So I have a fairly simple looking question I just forgot how to complete it.

2ydx = 3xdy when y(-2) = 1

dx/(3x) = dy(2y)

ln(3x) = ln(2y) + C

This is where I get stuck. I need to solve for y but I've just forgotten how to do this and don't know how to describe this type of problem.
Thanks for any help!
Exponentiate both sides and you get 3x = Ay where $A = 2e^C$ ....

3. Any chance you could walk through that with me? I'm afraid I don't know what you mean by exponentiate.

4. Originally Posted by coldize
Any chance you could walk through that with me? I'm afraid I don't know what you mean by exponentiate.
$e^{ln(3x)} = e^{ln(2y) + C}$ and apply the usual index and log rules.

5. Great. Thank you!