Differential Equations problem 2

Click For Summary
SUMMARY

The forum discussion revolves around solving the differential equation given by the expression 2(2x^2+y^2)dx-xydy=0. The user employs the substitution x = yv and derives a series of equations leading to the integral form. The critical step involves recognizing the need for partial fraction decomposition to solve the integrals. The user acknowledges a gap in their knowledge regarding partial fractions and expresses intent to review this concept for further progress.

PREREQUISITES
  • Understanding of differential equations and their solutions
  • Familiarity with substitution methods in differential equations
  • Knowledge of integration techniques, specifically partial fractions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review the method of partial fraction decomposition in calculus
  • Practice solving differential equations using substitution techniques
  • Explore integration techniques for complex rational functions
  • Study the application of differential equations in real-world scenarios
USEFUL FOR

Students studying differential equations, educators teaching calculus, and anyone looking to strengthen their integration and algebra skills in the context of differential equations.

Mastur
Messages
39
Reaction score
0

Homework Statement


2(2x^2+y^2)dx-xydy=0

Homework Equations


let x = yv

dx = ydv+vdy

The Attempt at a Solution


(4v^2y^2+2y^2)(ydv+vdy)-vy^2dy=0

4v^2y^3dv+4v^3y^2dy+2y^3dv+2vy^2dy-vy^2dy=0

4v^2y^3dv+2y^3dv+4v^3y^2dy+vy^2dy=0

2y^3(2v^2+1)dv+y^2(4v^3+v)dy=0
dividing all sides by y2
2y(2v^2+1)dv+(4v^3+v)dy=0
combining all ydys and vdvs
\frac{(2v^2+1)dv}{(4v^3+v)}+\frac{dy}{2y}=0

\int{\frac{(2v^2dv)}{4v^3+v}}+\int{\frac{dv}{4v^3+v}}+\int{\frac{dy}{2y}}=0

Err.. I don't know what to do next.. I can only integrate the 3rd integral..:frown:
 
Physics news on Phys.org
Two words. Partial fractions.
 
Naa, my memory.. I always forgot those!

Thank you very much for reminding! :biggrin:

I'll try to review first that partial fraction. It seems like I have forgotten the process of it already.
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
7
Views
4K
Replies
8
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K