# diff eq help

• Jul 12th 2007, 06:57 PM
suki0918
diff eq help
Find the particular solution of the differential equation: y" - 4y' + 4y = e^(2x)?

I'm not sure what to do b/c I get 0 for A?? Can someone help, this is what I had so far:

Y(x) = Ae^2x
Y'(x) = 2Ae^2x
Y"(x) = 4Ae^2x

So... 4Ae^2x - 8Ae^2x + 4Ae^2x = e^2x
Which becomes 0 = e^2x

and then I'm stumped...
• Jul 12th 2007, 07:29 PM
galactus
Have you ever used Variation of Parameters?.

$m^{2}-4m+4=0, \;\ (m-2)^{2}=0$

$y_{c}=C_{1}e^{2x}+C_{2}xe^{2x}$

$y_{1}=e^{2x}, \;\ y_{2}=xe^{2x}$

$W=\begin{vmatrix}e^{2x}&xe^{2x}\\2e^{2x}&2xe^{2x}+ e^{2x}\end{vmatrix}=e^{4x}$

$W_{1}=\begin{vmatrix}0&xe^{2x}\\e^{2x}&2xe^{2x}+e^ {2x}\end{vmatrix}=-xe^{4x}$

$W_{2}=\begin{vmatrix}e^{2x}&0\\2e^{2x}&e^{2x}\end{ vmatrix}=e^{4x}$

$u'_{1}=\frac{W_{1}}{W}=-x$

$u'_{2}=\frac{W_{2}}{W}=1$

$\int{-x}dx=\frac{-x^{2}}{2}$

$\int{1}dx=x$

$y_{p}=(\frac{-x^{2}}{2})e^{2x}+(x)xe^{2x}=\frac{x^{2}e^{2x}}{2}$

$y=y_{c}+y_{p}=C_{1}e^{2x}+C_{2}xe^{2x}+\frac{x^{2} e^{2x}}{2}$

$y=\left(\frac{x^{2}}{2}+C_{1}+C_{2}x\right)e^{2x}$

Whew, it's time for beddy-bye:D
• Jul 12th 2007, 08:13 PM
ThePerfectHacker
Quote:

Originally Posted by suki0918
Find the particular solution of the differential equation: y" - 4y' + 4y = e^(2x)?

Galactus used Variation of Parameters.

Another method is to use Method of Undetermined Coefficients.

The general solution is $y=C_1 e^{2x}+C_2xe^{2x}$

Since the RHS has $e^{2x}$ we look for a solution of the form $y=Ae^{2x}$. But that is among the solutions! So we multiply by $x$ to look for a solution of the form $y=Axe^{2x}$. But that is also among the solutions! So we look for a solution of the form $y=Ax^2e^{2x}$.
• Jul 12th 2007, 08:32 PM
servantes135
TPH,
Why didn't you use "Thou shall not covet anothers work as thine own"?
Lol verry funny though I must tell that to my Catholic friends.