Solving x^2 dy/dx = y-xy with y(-1)=-1

  • Thread starter Thread starter Saladsamurai
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on solving the differential equation x²(dy/dx) = y - xy with the initial condition y(-1) = -1. The user, Casey, initially derives the equation to ln(y) + C = -1/x - ln(x) but encounters issues with the initial values being outside the domain. Ben identifies a potential mistake in Casey's integration process, specifically regarding the application of the integral ∫(1/u) du = ln|u| + C, which leads to a resolution of the problem.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with integration techniques, particularly logarithmic integration
  • Knowledge of initial value problems in calculus
  • Concept of domain restrictions in mathematical functions
NEXT STEPS
  • Review techniques for solving first-order differential equations
  • Study the properties and applications of logarithmic functions in integration
  • Explore domain restrictions and their implications in calculus
  • Practice solving initial value problems with various differential equations
USEFUL FOR

Students and educators in calculus, particularly those focusing on differential equations and initial value problems, as well as anyone looking to refine their integration skills.

Saladsamurai
Messages
3,009
Reaction score
7

Homework Statement


Find implicit and explicit solution for

x^2\frac{dy}{dx}=y-xy when y(-1)=-1


So far I have:

\frac{(1-x)}{x^2}dx=dy/y

\Rightarrow x^{-2}(1-x)dx=\ln y+C

\Rightarrow (x^{-2)-x^{-1})dx=\ln y+C

\Rightarrow -\frac{1}{x}-\ln x=\ln y+C

With the initial values, (-1,-1) are outside of the domain. What gives? My integration looks good to me. What am I missing?

Thanks,
Casey
 
Physics news on Phys.org
Actually,

\int \frac{du}{u} = \ln |u| + C

Assuming everything else is correct, that might be your mistake.
 
Oh, yes, that looks like it would clear up the problem. Thanks Ben. The integration looked a little to easy to be getting the best of me. It's always the details...

Thanks,
Casey
 

Similar threads

Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
14
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K