Seperation of Variables/Integrating Factor Method

  • Thread starter _N3WTON_
  • Start date
  • #1
351
3

Homework Statement


The first order linear equation of the form:
[itex]\frac{dy}{dx} + ay = b [/itex]
where a and b are constants, can be solved both by the integrating factor method and by seperation of variables. Solve the equation using both methods to see that you get the same solution.



Homework Equations


Separation of variables and Integrating factor equations

The Attempt at a Solution


I am 80% sure that I have correctly found the solution using both methods; however, I am having trouble equating the two solutions. This is what I have done so far, first I will show using the integrating factor method, followed by separation of variables:
[itex] p(x) = a [/itex]
[itex] u(x) = e^{ax} [/itex]
[itex] \frac{d}{dx}(e^{ax} y(x)) = be^{ax} [/itex]
[itex] \int\frac{d}{dx}(e^{ax} y(x))\,dx = \int be^{ax}\, dx [/itex]
[itex] e^{ax} y(x) = abe^{ax} + C [/itex]
[itex] y(x) = ab + \frac{C}{e^ax}[/itex]
Now again using the separation of variables method:
[itex] \frac{dy}{dx} = b-ay [/itex]
[itex] \frac{dy}{y} = (b-a)dx [/itex]
[itex] \int\frac{dy}{y} = \int b-a\, dx [/itex]
[itex] ln(y) = bx - ax + C [/itex]
[itex] y = e^{bx} - e^{ax} + e^C [/itex]
 

Answers and Replies

  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement


The first order linear equation of the form:
[itex]\frac{dy}{dx} + ay = b [/itex]
where a and b are constants, can be solved both by the integrating factor method and by seperation of variables. Solve the equation using both methods to see that you get the same solution.



Homework Equations


Separation of variables and Integrating factor equations

The Attempt at a Solution


I am 80% sure that I have correctly found the solution using both methods; however, I am having trouble equating the two solutions. This is what I have done so far, first I will show using the integrating factor method, followed by separation of variables:
[itex] p(x) = a [/itex]
[itex] u(x) = e^{ax} [/itex]
[itex] \frac{d}{dx}(e^{ax} y(x)) = be^{ax} [/itex]
[itex] \int\frac{d}{dx}(e^{ax} y(x))\,dx = \int be^{ax}\, dx [/itex]
[itex] e^{ax} y(x) = abe^{ax} + C [/itex]
[itex] y(x) = ab + \frac{C}{e^ax}[/itex]
Now again using the separation of variables method:
[itex] \frac{dy}{dx} = b-ay [/itex]
[itex] \frac{dy}{y} = (b-a)dx [/itex]
[itex] \int\frac{dy}{y} = \int b-a\, dx [/itex]
[itex] ln(y) = bx - ax + C [/itex]
[itex] y = e^{bx} - e^{ax} + e^C [/itex]

Never write ##1/e^{ax}##; always write it as ##e^{-ax}##, because that way you can apply standard differentiation/integration formulas.

Your second method has an obvious blunder: from
[tex] \frac{dy}{dx} = b-ay [/tex]
it dos NOT follow that
[tex] \frac{dy}{y} = (b-a)dx \: \longleftarrow \:\text{false}[/tex]
 
  • #4
351
3
Never write ##1/e^{ax}##; always write it as ##e^{-ax}##, because that way you can apply standard differentiation/integration formulas.

Your second method has an obvious blunder: from
[tex] \frac{dy}{dx} = b-ay [/tex]
it dos NOT follow that
[tex] \frac{dy}{y} = (b-a)dx \: \longleftarrow \:\text{false}[/tex]
im a little confused about what is wrong with the separation I performed?
Edit: Nvm I see it now, thank you
 
Last edited:

Related Threads on Seperation of Variables/Integrating Factor Method

  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
18
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
1
Views
1K
Replies
7
Views
2K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
6
Views
814
Replies
5
Views
2K
Replies
7
Views
2K
Top