# Linear Differential Equations

Need some hints or help with this problem:

Find complete solution for $$u^{'''}-6u^{''}+9u^{'}=4e^{-t}$$ and solve the initial value problem with initial values as $$u(0)=\frac{1}{2}, u^{'}(0)=\frac{11}{4}, u^{''}(0)=\frac{41}{4}$$

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Laplace transform allowed? – user13838 Oct 26 '11 at 16:44
What have you done so far, and where are you stuck? – Olivier Bégassat Oct 26 '11 at 16:51

## 3 Answers

First you change variables: $y = u^\prime$, so your equation becomes

$y^{\prime\prime} - 6 y^\prime + 9 y = 4 e^{-t}$

This is a 2nd order equation with constant coefficients. The homogeneous part of the equation has the solution

$y_h = A e^{3t} + B t e^{3t}$,

with $A$ and $B$ arbitrary constants. You can guess that the particular solution of the inhomogeneus part will presumably be proportional to $e^{-t}$ (which it is). You can easily find that

$y_p = \displaystyle\frac{e^{-t}}{4}$.

Therefore, the total solution for $y$ is

$y = y_h + y_p = A e^{3t} + B t e^{3t} + \displaystyle\frac{e^{-t}}{4}$.

To find $u$, you just find the primitive of this solution, which is rather standard:

$u = \alpha e^{3t} + \beta t e^{3t} - \displaystyle\frac{e^{-t}}{4} + \gamma$,

where $\alpha$, $\beta$ and $\gamma$ are other arbitrary constants that can be determined using the initial conditions. I stop here, I think the rest is the easy part.

Cheers.

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Thank you, managed to solve the problem. – krunarsson Oct 27 '11 at 14:47

Show that $u_0(t)=\frac{-e^{-t}}{4}$ is a solution (without the conditions.) Then in general $u(t)$ is a solution if and only if $v(t)=u(t)-u_0(t)$ is a solution of $v^{'''}-6v^{''}+9v^{'}=0$ Alter your initial conditions on $u$ to be conditions on $v$.

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Total solution of LDE is given by:

$u=u_h+u_p$

Since characteristic equation is:

$r^3-6r^2+9r=0\Rightarrow r(r-3)^2=0\Rightarrow$

$\Rightarrow u_h=C_1+C_2e^{3t}+C_3te^{3t}$

In order to find particular solution we can use method of variation of parameters.Particular solution is of the form:

$u_p=a(t)+b(t)e^{3t}+c(t)te^{3t}$

We can find $a(t),b(t),c(t)$ if we solve following system:

$\begin{cases} a'+b'e^{3t}+c'te^{3t}=0 \\ 3b'e^{3t}+(e^{3t}+3te^{3t})c'=0 \\ 9b'e^{3t}+(6e^{3t}+9te^{3t})c'=4e^{-t} \end{cases}$

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