Differential Equation- Separation of Variables

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Homework Help Overview

The discussion revolves around solving a differential equation using the method of separation of variables, specifically the equation y' = x sec²(x²). Participants are exploring the integration process and substitution techniques involved in the solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of u-substitution for integration, with one suggesting a substitution of u = x² and questioning the complexity of the resulting integral. Others propose rewriting the sec function in terms of cosine to simplify the integration process.

Discussion Status

The discussion is active, with various participants offering different substitution methods and corrections to earlier attempts. There is a focus on clarifying the correct form of the integrals and ensuring the substitutions are accurate, but no consensus has been reached on a final approach.

Contextual Notes

Some participants note potential errors in the original substitution and integration steps, highlighting the importance of careful manipulation of the differential equation. There is an acknowledgment of the challenges posed by the integration process.

brutalmadness
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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
\intdy=\intxsec^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?
 
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brutalmadness said:
lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx
This is incorrect. Your method is correct, though.
 
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.
 
u=x^2 du=2xdx
2\intdu/cos^2(u)
 
You're off by a factor of 1/4. It should be \frac{1}{2} \int sec^2u \ du.
 

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