- #1

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## 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