Please help me solve this differential equation for the initial condition (0,-1)

Click For Summary
SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) given by dx/dy = ((1+x^2)^(1/2))/(xy^3) with the initial condition (0,-1). The user encounters an issue where substituting the initial condition leads to the square root of negative one, indicating a potential error in their integration process. The correct approach involves separating variables and integrating to obtain the expression ((x^2+1)^(1/2)) = -1/2 * y^(-2) + C, which must be evaluated carefully to satisfy the initial condition.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Knowledge of variable separation technique in calculus
  • Familiarity with integration methods
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the method of separation of variables in ODEs
  • Learn about integrating factors for solving first-order ODEs
  • Explore the implications of initial conditions in differential equations
  • Investigate complex numbers and their role in solving equations involving square roots of negative values
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations, as well as educators seeking to clarify integration techniques and initial condition applications.

hornekat
Messages
1
Reaction score
0
Please help me solve this differential equation for the initial condition (0,-1):
dx/dy = ((1+x^2)^(1/2))/(xy^3)

I think I'm doing something wrong because I end up with
((x^2)(y^3))/2 = ((x^2)+y)^(1/2) + c,
but when plugging in the initial condition it ends up being the square root of negative one.

Please help, thank you!
 
Physics news on Phys.org
The ODE is:

$$\d{x}{y}=\frac{\left(x^2+1\right)^{\frac{1}{2}}}{xy^3}$$

Separating variables, we obtain:

$$\frac{x}{\left(x^2+1\right)^{\frac{1}{2}}}\,dx=y^{-3}\,dy$$

And now integrating, we get:

$$\left(x^2+1\right)^{\frac{1}{2}}=-\frac{1}{2}y^{-2}+C$$
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
2K
  • · Replies 52 ·
2
Replies
52
Views
8K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K