2nd order differential equation question

[NEGATIVE_40]

Homework Statement

$$y''-3y'=4$$

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:
$$y''-3y'=4$$
therefore $$r^2-3r=0$$ so $$r =0,~3$$
giving a complementary solution of $$y_{c} = c_{1} +c_{2}e^{3x}$$

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

So the particular solution is therefore $$0 - 3 \cdot 4 = 4$$ giving $$-12 = 4$$

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

 

[vela]

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.

 

[NEGATIVE_40]

so ... my first particular solution would be $$y_p = Ax$$ ?
If I do this I get $$y_p = - \frac{4}{3}x$$

so finally I end up with $$y= c_1 + c_{2}e^{3x} -\frac{4}{3}x$$
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 $$c_1$$ being a part of the complementary solution (the r=0 part) , and you could mulitply this by some number to get A?

 

[vela]

I think you got it, but just to be clear, let me reword it:

You have two complementary solutions, y₁=e^{0x[/sub]=1 and y₂=e3x. If you tried to use y_p=A, it wouldn't be linearly independent of the complementary solutions because y_p=Ay₁.}

 

[NEGATIVE_40]

that all makes sense. thanks for the help vela.