Differential Equation- Separation of Variables

[brutalmadness]

Homework Statement

y'=xsec^2(x^2)

2. The attempt at a solution
dy/dx=xsec^2(x^2)
dy=xsec^2(x^2)dx
[tex]\int[/tex]dy=[tex]\int[/tex]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?

 

[Defennder]

lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx

This is incorrect. Your method is correct, though.

 

[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.

 

[brutalmadness]

u=x^2 du=2xdx
2[tex]\int[/tex]du/cos^2(u)

 

[Defennder]

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