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[1] $y'' - 4y' + 4y = 0$

Usually problem like these will have the answer in the form $C_1e^a + C_2e^b ... $ where $a $ and $b$ are the roots of the characteristic equation $e^{rt}$

$$ y = e^{rt} $$ $$ y' = re^{rt}.. y'' = r^2 e^{rt} $$ $$ r^2e^{rt} - 4e^{rt} + 4e^{rt} = 0$$ $$ e^{rt}(r-2)(r-2) = 0$$ $$ y= C_1e^{2t} + C_2e^{2t}$$

However, this is not correct as the answer is $ y = C_1e^{2t} + C_2te^{2t}$ ! I just don't know why.

[2] For a similar problem , $y'' + 3y' - 4y = 0 $ I did the exact same thing and the answer is $$y = c_1e^t + c_2e^{-4t} $$

How are [1] and [2] different? They look the same, why does [2]'s solution have an extra factor of t.

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It's because the characteristic polynomial in [1] has a repeated root. See here. –  Clive Newstead Feb 17 '13 at 23:43

2 Answers 2

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The difference is that [1] has a repeated root. When you have a root $r$ that is repeated $n$ times, its contribution to the solution is of the form:

$$ (c_0 + c_1 t + \cdots + c_n t^n) e^{r t} $$

When $n = 1$, this becomes:

$$ (c_0 + c_1 t) e^{r t} $$

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is it considered to have "repeated roots" if the roots are (r-2)(r+2)? (roots of 2 and -2). No right? –  40Plot Feb 17 '13 at 23:49
    
@40Plot: No, because $2 \ne -2$. –  Clive Newstead Feb 17 '13 at 23:51
    
@40Plot As Clive Newstead points out, $2$ and $-2$ are different roots. If $2$ is repeated twice, you'll get $(r-2)^2$. –  Ayman Hourieh Feb 17 '13 at 23:55

Consider $\rm D$ as the derivation operator on the space of functions you want to study. This just says that $\mathrm D f = f'$.

Then your differential equation can be written as "Find the kernel of the linear operator $\rm D^2 - 4 \rm D + 4$". The theory of differential equation tells you that this kernel has dimension 2.

Let's find a basis of it, $\rm D^2 - 4 D + 4$ can be rewritten as $(\rm D - 2)^2$, so if I know a basis of the kernel of $\rm D - 2$ it will be part of a basis we want to find. Here it is the function $x \mapsto e^{2x}$. But this is not enough, because the kernel of $\rm D -2$ has only dimension 1. Then we need to add $x \to x e^{2x}$ which is not leaving in the kernel of $\rm D - 2$.

To compare with your second example, here you want to find the kernel of $\rm D^2 + 3 D - 4$ which is $(\rm D - 1)(D + 4)$. Now, as before, the kernel has dimension 2, but as the characteristic equation has two distinct roots, you can take a vector from the basis of $\rm D -1$ and complete with another vector of the basis of the kernel of $\rm D + 4$.

In conclusion : it all depends on whether the characteristic equation has multiple roots or not.

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Then in what case will it has a form of $Ce^tcos(t)...Ce^tsin(t) $ –  40Plot Feb 18 '13 at 0:17

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