First order linear DEs

  • Thread starter BOAS
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  • #1
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Hi,

I have a first order linear DE that I need to find the general solution for. I thought that I had, but my solution does not make sense when plugged back into the equation.

I think that my method of separation of variables might be inapplicable here, but don't know the reason for this.

I know that for a DE of the form;

[itex]y' + ay = b[/itex], the general solution is given by [itex]y = \frac{b}{a} + Ce^{-ax}[/itex] but I don't know where this result comes from, so I would greatly appreciate some help here.

I will show what I have done, even though it is wrong.

1. Homework Statement


Find the general solution to;

[itex]c(\phi) : c' + 2c = 1[/itex]

Homework Equations




The Attempt at a Solution


[/B]
Rewrite as [itex]\frac{dc}{d \phi} = 1 - 2c[/itex]

[itex]\frac{dc}{1 - 2c} = d\phi[/itex]

Integrate both sides.

[itex]\int \frac{dc}{1 - 2c} = \int d\phi[/itex]

[itex]- \frac{1}{2} ln(|1 - 2c|) + a_{1} = \phi + a_{2}[/itex]

Exponentiate, to get rid of ln, and let a_2 - a_1 = a

[itex](1 - 2c)^{-\frac{1}{2}} = e^{\phi} + e^{a}[/itex]

[itex]c = \frac{1}{2} - \frac{e^{-2\phi} + e^{-2a}}{2}[/itex]

which is not a solution to my differential equation...

[itex]c = \frac{1}{2} + Qe^{-2 \phi}[/itex] is. (Using Q as my constant to avoid confusion)

So I have two questions;

1) Why can I not use separation of variables? (Or did I make a mistake?)

2) What is the argument that leads to the aforementioned 'formula' for solutions?

Thanks.
 
Last edited:

Answers and Replies

  • #2
944
394
Your left side antiderivative is incorrect, and you didn't take the exponential of the entire right side.

The general way of solving nth order linear differential equations with constant coefficients (which really show up everywhere) is to assume a solution of [itex]e^{st}[/itex], plug it in to the ODE in the homogeneous case, and solve for [itex]s[/itex]. If the forcing function is not 0, then there are a couple of techniques to find the remaining term (since the sum of solutions to an ODE is a solution). For a constant, it's easy enough to just assume a constant solution and find out what it's equal to.
 
  • #3
Ray Vickson
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Hi,

I have a first order linear DE that I need to find the general solution for. I thought that I had, but my solution does not make sense when plugged back into the equation.

I think that my method of separation of variables might be inapplicable here, but don't know the reason for this.

I know that for a DE of the form;

[itex]y' + ay = b[/itex], the general solution is given by [itex]y = \frac{b}{a} + Ce^{-ax}[/itex] but I don't know where this result comes from, so I would greatly appreciate some help here.

I will show what I have done, even though it is wrong.

1. Homework Statement


Find the general solution to;

[itex]c(\phi) : c' + 2c = 1[/itex]

Homework Equations




The Attempt at a Solution


[/B]
Rewrite as [itex]\frac{dc}{d \phi} = 1 - 2c[/itex]

[itex]\frac{dc}{1 - 2c} = d\phi[/itex]

Integrate both sides.

[itex]\int \frac{dc}{1 - 2c} = \int d\phi[/itex]

[itex]- \frac{1}{2} ln(|1 - 2c|) + a_{1} = \phi + a_{2}[/itex]

Exponentiate, to get rid of ln, and let a_2 - a_1 = a

[itex](1 - 2c)^{-\frac{1}{2}} = e^{\phi} + e^{a}[/itex]

************************************************************** ERROR above

[itex]c = \frac{1}{2} - \frac{e^{-2\phi} + e^{-2a}}{2}[/itex]

which is not a solution to my differential equation...

[itex]c = \frac{1}{2} + Qe^{-2 \phi}[/itex] is. (Using Q as my constant to avoid confusion)

So I have two questions;

1) Why can I not use separation of variables? (Or did I make a mistake?)

2) What is the argument that leads to the aforementioned 'formula' for solutions?

Thanks.

##e^{a + \phi} \neq e^a + e^{\phi}##.

In general, for constants ##a,b## we can solve ## y' + ay = b##, by setting ##z = y - a/b##. Then ##z' = y'##, and so ##z' + az = y' + a(y - b/a) = y' + ay - b = 0##; that is, ##z' + az = 0##.
 
  • #4
117
7
Also, you can use an integrating factor: [tex]let\, \mu =e^{\int p(\phi )d\phi }[/tex]

muliply your equation through by [tex]\mu[/tex] so you have

[tex]\mu c'+2\mu c=\mu[/tex] then remind yourself of the chain rule. The answer is only two steps away from here... hint: [tex]\mu c' +2\mu c =(\mu c)' [/tex]
 
  • #5
555
19
##e^{a + \phi} \neq e^a + e^{\phi}##.

Oops, silly mistake.

##(1 - 2c)^{-\frac{1}{2}} = e^{\phi + a}## Which can be rewritten as ##(1 - 2c)^{-\frac{1}{2}} = e^{\phi} e^{a}##

Let ##A = e^{2a}##

##c = \frac{1}{2} - 2A^{-1} e^{-2\phi}## Which is a solution.

n general, for constants a,b we can solve y′+ay=b, by setting z=y−a/b. Then z′=y′, and so z′+az=y′+a(y−b/a)=y′+ay−b=0; that is, z′+az=0.

Thank you for your help.

Practice makes perfect, so I will do lots of problems similar to this.
 

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