# Differential Equation- Separation of Variables

1. May 19, 2008

1. The problem statement, all variables and given/known data
y'=xsec^2(x^2)

2. The attempt at a solution
dy/dx=xsec^2(x^2)
dy=xsec^2(x^2)dx
$$\int$$dy=$$\int$$xsec^2(x^2)dx
lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx

... and here's my problem. It seems like that creates a very difficult u-sub to try and manage. Any suggestions/help?

2. May 19, 2008

### Defennder

This is incorrect. Your method is correct, though.

3. May 19, 2008

### soul

I think you should write sec function in terms of cos. Function will be dy = x / (cos^2(x^2)). Then if you say x^2 = u and du = 2x.dx, you can solve the problem easily.

4. May 19, 2008

u=x^2 du=2xdx
2$$\int$$du/cos^2(u)

5. May 19, 2008

### Defennder

You're off by a factor of 1/4. It should be $$\frac{1}{2} \int sec^2u \ du$$.