Solving First Order ODE with Integrating Factor

In summary, the conversation is about solving a first order ODE using an integrating factor. The attempt at a solution involves finding the integrating factor and using it to solve the equation. However, there is a mistake in the solution where the "1" term is dropped, leading to a different answer than what was obtained using Maple.
  • #1
bobred
173
0

Homework Statement


Solve first order ODE

Homework Equations



[tex]\frac{dy}{dx}=x^2+1+\frac{2}{x}y[/tex]
Rearranged
[tex]\frac{dy}{dx}-\frac{2}{x}y=x^2+1[/tex]

The Attempt at a Solution


Integrating factor
[tex]p=\exp(-\int \frac{2}{x})=\exp(-2\ln x)=x^{-2}[/tex]

Multiplying through by the integrating factor
[tex]\frac{d}{dy}(x^{-2}y)=x^{-2}[/tex]

Integrating both sides
[tex]x^{-2}y=-x^{-1}+C[/tex]

Dividing through by [tex]x^{-2}[/tex]
[tex]y=Cx^2-x[/tex]

The problem comes when I use say, Maple to check the answer, it gives

[tex]y=x^3+Cx^2-x[/tex]

Any ideas? Thanks
 
Physics news on Phys.org
  • #2
bobred said:

Homework Statement


Solve first order ODE


Homework Equations



[tex]\frac{dy}{dx}=x^2+1+\frac{2}{x}y[/tex]
Rearranged
[tex]\frac{dy}{dx}-\frac{2}{x}y=x^2+1[/tex]

The Attempt at a Solution


Integrating factor
[tex]p=\exp(-\int \frac{2}{x})=\exp(-2\ln x)=x^{-2}[/tex]

Multiplying through by the integrating factor
[tex]\frac{d}{dy}(x^{-2}y)=x^{-2}[/tex]
No, the right hand side of your original equation was [itex]x^2+ 1[/itex]. Multiplying that by [itex]x^{-2}[/itex] gives [itex]1+ x^{-2}[/itex]. You've dropped the "1".

Integrating both sides
[tex]x^{-2}y=-x^{-1}+C[/tex]

Dividing through by [tex]x^{-2}[/tex]
[tex]y=Cx^2-x[/tex]

The problem comes when I use say, Maple to check the answer, it gives

[tex]y=x^3+Cx^2-x[/tex]

Any ideas? Thanks
 
  • #3
Thanks
 

1. What is a first order ODE?

A first order ordinary differential equation (ODE) is an equation that involves only one independent variable and its derivatives. It can be written in the form: F(x,y,y') = 0, where x is the independent variable, y is the dependent variable, and y' is the derivative of y with respect to x.

2. What is the general solution to a first order ODE?

The general solution to a first order ODE is the most general form of the solution that satisfies the given equation. It typically contains one or more arbitrary constants, which can be determined by applying initial or boundary conditions.

3. How do you solve a first order ODE analytically?

To solve a first order ODE analytically, you can use various methods such as separation of variables, integrating factor, or substitution. These methods involve manipulating the equation to isolate the dependent and independent variables, and then integrating both sides to find the solution.

4. What is the difference between an explicit and implicit solution to a first order ODE?

An explicit solution to a first order ODE is one that can be written explicitly in terms of the independent variable and constants, while an implicit solution cannot be written explicitly and may involve an implicit relationship between the variables.

5. Can numerical methods be used to solve first order ODEs?

Yes, numerical methods can be used to solve first order ODEs. These methods involve approximating the solution through a series of calculations, and can be more efficient for complex or non-analytical equations. Examples of numerical methods include Euler's method, Runge-Kutta methods, and finite difference methods.

Similar threads

  • Calculus and Beyond Homework Help
Replies
19
Views
765
  • Calculus and Beyond Homework Help
Replies
8
Views
752
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
442
  • Calculus and Beyond Homework Help
Replies
5
Views
616
  • Calculus and Beyond Homework Help
Replies
14
Views
201
  • Calculus and Beyond Homework Help
Replies
4
Views
485
  • Calculus and Beyond Homework Help
Replies
21
Views
823
  • Calculus and Beyond Homework Help
Replies
10
Views
418
  • Calculus and Beyond Homework Help
Replies
6
Views
746
Back
Top