Substitution to convert first order ODE to homogenous

Click For Summary
SUMMARY

The discussion focuses on transforming the first-order ordinary differential equation (ODE) given by ##\frac{dy}{dx}=\frac{2x+y-3}{x-2y+1}## into a homogeneous equation using the substitutions ##x=X+h## and ##y=Y+k##. The resulting homogeneous equation is ##\frac{dY}{dX}=\frac{2X+Y}{X-2Y}##. The key challenge addressed is understanding how the derivative transformation from ##\frac{dy}{dx}## to ##\frac{dY}{dX}## occurs, particularly through the application of the chain rule. The linear nature of the substitutions is emphasized as crucial for this transformation.

PREREQUISITES
  • Understanding of first-order ordinary differential equations (ODEs)
  • Familiarity with substitution methods in differential equations
  • Knowledge of the chain rule in calculus
  • Basic concepts of homogeneous functions and equations
NEXT STEPS
  • Study the method of substitution in solving first-order ODEs
  • Learn about homogeneous equations and their properties
  • Explore the application of the chain rule in calculus with examples
  • Investigate linear transformations in differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as anyone seeking to deepen their understanding of ODE transformations and substitution techniques.

stfz
Messages
35
Reaction score
0

Homework Statement


Use the substitution ##x=X+h## and ##y=Y+k## to transform the equation
##\frac{dy}{dx}=\frac{2x+y-3}{x-2y+1}## to the homogenous equation
##\frac{dY}{dX}=\frac{2X+Y}{X-2Y}##
Find h and k and then solve the given equation

Homework Equations

The Attempt at a Solution


If I simply make the substitution into the equation, I get a homogenous equation which I can solve using y=vx substitution. But what I need help understanding is how the ##\frac{dy}{dx}## becomes ##\frac{dY}{dX}## after simply substituting into the LHS?
Is some proof or method of doing this so that I can turn dy/dx into dY/dX and vice versa? The chain rule doesn't help, as I cannot relate X and Y
 
Physics news on Phys.org
First \frac{dY}{dx}= \frac{d(y- k)}{dx}= \frac{dy}{dx}

Now use the chain rule. \frac{dY}{dX}= \frac{dY}{dx}\frac{dx}{dX}= \frac{dy}{dx}(1)

The real point is that both x= X+ h and y= Y+ k are linear with slope 1.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
19
Views
3K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
11
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
Replies
4
Views
2K