Integrating Factor for Solving ODE with Linear Coefficients

In summary, the conversation is about solving the problem xy'+2y=3x by dividing by x and finding the integrating factor e^{\int \frac{2}{x}dx}. The correct integrating factor is x^2, not 2x. Multiplying the equation by x^2 and taking the integral/derivative of the integrating factor leads to the solution y=\frac{x^{3}+C}{x^{2}}.
  • #1
iRaid
559
8
Problem:
[tex]xy'+2y=3x[/tex]
Attempt:
Divide by x...
[tex]y'+\frac{2y}{x}=3[/tex]
I think I find the integrating factor by doing:
[tex]e^{\int \frac{2}{x}dx}[/tex]

Not sure if that's right but if it is then the solution to the integral is just 2x.

Any help is appreciated
 
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  • #2
You will need to add in the constant of integration to the right side before you do any simplification.
 
  • #3
iRaid said:
Problem:
[tex]xy'+2y=3x[/tex]
Attempt:
Divide by x...
[tex]y'+\frac{2y}{x}=3[/tex]
I think I find the integrating factor by doing:
[tex]e^{\int \frac{2}{x}dx}[/tex]

Not sure if that's right but if it is then the solution to the integral is just 2x.

Any help is appreciated
Yes, the integrating factor is [itex]e^{\int (2/x)dx}[/itex] but then the integral is 2 ln(x) so the integrating factor is [itex]e^{2ln(x)}= e^{ln(x^2)}= x^2[/itex], not 2x.

Multiplying the equation by that gives [itex]x^2y'+ 2xy= (x^2y)'= 3x^2[/itex]
 
Last edited by a moderator:
  • #4
Yeah I figured it out ends up:
[tex]x^{2}y'+2xy=3x^{2}[/tex]
Take integral/derivative of integrating factor:
[tex]\int \frac{d}{dx} x^{2}y=\int 3x^{2}dx[/tex]
Simplifies to:
[tex]x^{2}y=x^{3}+C \implies y=\frac{x^{3}+C}{x^{2}}[/tex]
(can be simplified a little more too)

Thanks for the help.
 

1. What is an ODE with integrating factor?

An ODE with integrating factor is a type of ordinary differential equation (ODE) that involves the use of an integrating factor to simplify the equation and make it easier to solve. The integrating factor is a function that is multiplied to both sides of the equation to convert it into a more manageable form.

2. Why do we use an integrating factor in ODEs?

An integrating factor is used in ODEs to make the equation easier to solve by reducing the complexity of the equation. It also helps in finding a general solution to the ODE, rather than a specific solution for a particular set of initial conditions.

3. How do you find the integrating factor in an ODE?

The integrating factor in an ODE can be found by using the formula e∫P(x)dx, where P(x) is the coefficient of the highest derivative term in the equation. This formula can also be extended to include more complicated functions, such as trigonometric or logarithmic functions.

4. Can we always use an integrating factor to solve an ODE?

No, not all ODEs can be solved using an integrating factor. It depends on the form of the equation and the types of functions involved. In some cases, other methods such as separation of variables or substitution may be more suitable for solving the ODE.

5. What are some real-world applications of ODEs with integrating factor?

ODEs with integrating factor have many real-world applications, including in physics, biology, economics, and engineering. For example, they can be used to model the spread of diseases, the growth of populations, and the movement of fluids in pipes. They are also used in circuit analysis and in calculating the deflection of beams under load.

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