Question

Solve the differential equation dy/dx = y/(2x^2) with the initial condition of f(2) = 1.

Answer 1

\[\Large \frac{dy}{dx}=\frac{y}{2x^2} \]

Separation of variables, can you try doing that?

Answer 2

I have that bit:\[\int\limits_{?}^{?} 2dy/y = \int\limits_{?}^{?} dx/x^2\]

Answer 3

yes

Answer 4

I'm unsure where to continue from there. Actually integrating them, especially the right side, poses the problem.

Answer 5

It's seperable so you have\[\int\limits_{1}^{y}\frac{du}{u} = \int\limits_{2}^{x}\frac{dv}{2v^{2}}\]

Answer 6

@sask54, I understand, but it's really important for Differential Equations that you feel absolutely comfortable with computing integrals.

However, often it helps to rewrite your problem:

\[\Large 2\int\frac{1}{y}dy=\int x^{-2}dx \]

Answer 7

Oh, I completely missed that bit. I kept getting variations resulting in e's. Sorry for the silly mistake. Thanks!

Answer 8

You're welcome!