2nd order differential equation question

  • #1

Homework Statement



[tex] y''-3y'=4 [/tex]

Homework Equations



none

The Attempt at a Solution


I've been looking at this problem for the last hour and have absolutely no idea on how to solve it.

My initial approach is this:
[tex] y''-3y'=4 [/tex]
therefore [tex] r^2-3r=0 [/tex] so [tex] r =0,~3[/tex]
giving a complementary solution of [tex]y_{c} = c_{1} +c_{2}e^{3x} [/tex]

Now this is the part where I think I might be making a mistake
[tex] y_{p} = 4A [/tex] (so I get any multiple of 4)
[tex] y_{p}' = 4 [/tex]
[tex] y_{p}'' = 0 [/tex]

So the particular solution is therefore [tex] 0 - 3 \cdot 4 = 4 [/tex] giving [tex] -12 = 4 [/tex]

I can't think of another way to this, so any help would be appreciated :smile:
 

Answers and Replies

  • #2
vela
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The problem is that your particular solution is a multiple of one of your complementary solutions, namely the r=0 solution. You need to modify your trial solution to make it linearly independent of the other solutions.
 
  • #3
so ... my first particular solution would be [tex] y_p = Ax [/tex] ?
If I do this I get [tex] y_p = - \frac{4}{3}x [/tex]

so finally I end up with [tex] y= c_1 + c_{2}e^{3x} -\frac{4}{3}x [/tex]
which is the correct answer.

I think I understand what you said about the particular solution being a multiple of the complementary solution, but just to clarify.

Is this what you mean: I got [tex] c_1 [/tex] being a part of the complementary solution (the r=0 part) , and you could mulitply this by some number to get A?
 
Last edited:
  • #4
vela
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I think you got it, but just to be clear, let me reword it:

You have two complementary solutions, y1=e0x[/sub]=1 and y2=e3x. If you tried to use yp=A, it wouldn't be linearly independent of the complementary solutions because yp=Ay1.
 
  • #5
that all makes sense. thanks for the help vela.
 

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