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Differential Equation- Separation of Variables

  • #1

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?
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx
This is incorrect. Your method is correct, though.
 
  • #3
62
0
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
u=x^2 du=2xdx
2[tex]\int[/tex]du/cos^2(u)
 
  • #5
Defennder
Homework Helper
2,591
5
You're off by a factor of 1/4. It should be [tex]\frac{1}{2} \int sec^2u \ du[/tex].
 

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