# Initial-Value Problem Integration Question

1. Aug 29, 2011

### bobbles22

Hi there,

I'm doing a maths uni pre course with some questions on differentiation and integration.

I'm looking for a few pointers on how to proceed.

I need to solve:

dy/dx=(cos(3x))/(2-sin(3x)) when y=2 and x=0

Can anyone help with how I'm meant to approach this and how I go about actually integrating the equation. Its been some time since I've done things like this and the information I have is not very easy to understand.

Many thanks

Bob

2. Aug 29, 2011

### jackmell

you got:

$$\frac{dy}{dx}=\frac{\cos(3x)}{2-\sin(3x)}$$

You can now separate variables and integrate:

$$\int_{y_0}^y dy=\int_{x_0}^x \frac{\cos(3x)}{2-\sin(3x)}dx$$

The left side is just y-y_0. The other side you can do right? Just make a u-substitution. For example, what is the derivative of the denominator? Isn't that real close to what's in the numerator? Can you adjust it to be just right then integrate?