# Find the solution of the given initial value problem $y^{(4)}-4y'''+4y''=0$

I can't really figure this out but this is what I've done some far:

$y^{(4)}-4y'''+4y''=0$ where $y(1)=9+e^2$ $y'(1)=7+2e^3$ $y''(1)=4e^2$ and $y'''(1)=8e^2$

I found the characteristic equation $r^4-4r^3+4r^2=0$ and it factored to $r^2(r-2)^2=0$

So $y=c_1+c_2+c_3e^{2t}+c_4te^{2t}$ and then found the derivatives and plugged in the initial values to get

$9+e^2=c_1+c_2+c_3e^2+c_4e^2$

$7+2e^2=2c_3e^2+3c_4e^2$

$4e^2=4c_3e^2+8c_4e^2$

$8e^2=8c_3e^2+20c_4e^2$

But when I solve, I get $c_3=1$ and $c_4=0$ but thats not right. What am I doing wrong?

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Now it's been a while since I've been doing this, but shouldnt't it be $y=c_1+c_{2}t+c_{3}e^{2t}+c_{4}te^{2t}$?

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Yes, the way to handle repeated roots is to just multiply by a $t$. If $(\frac{d}{dt} - r)^2 y = 0$, then $(\frac{d}{dt} - r) z = 0$ where $z = (\frac{d}{dt} - r) y$. Solving this gives us $z = c_1 e^{rt}$ and so $(\frac{d}{dt} - r) y = c_1 e^{rt}$. Using an integrating factor, we see that $y = c_1 t e^{rt} + c_2 e^{rt}$. This same process shows how to handle higher order repeated roots too. – muzzlator Mar 5 '13 at 1:14

Essentially a typo: it should be $c_1+c_2 t$, not $c_1+c_2$. That makes the second linear equation you got incorrect. (The first is right because $t$ happens to be $1$.)

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