Okay.. Can anyone explain me, how to solve this problem:

$\displaystyle y''+4y'+4=0$.. when $\displaystyle y(0)=2$ and $\displaystyle y'(0)=0$

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- Sep 25th 2009, 08:46 AM #1

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- Sep 25th 2009, 08:57 AM #2
The Characteristic Equation is

$\displaystyle m^2 + 4m + 4 = 0$

$\displaystyle (m + 2)^2 = 0$

$\displaystyle m = -2$.

Since the solution of the characteristic equation is repeated, the solution will be of the form

$\displaystyle y(x) = C_1e^{-2x} + C_2xe^{-2x}$.

We can also see that

$\displaystyle y'(x) = -2C_1e^{-2x} + C_2e^{-2x} -2C_2xe^{-2x}$.

Applying the initial conditions gives

$\displaystyle 2 = C_1 + C_2$ and $\displaystyle 0 = -2C_1 + C_2$.

Solving the equations simultaneously gives

$\displaystyle 2 = 3C_1$

$\displaystyle \frac{2}{3} = C_1$

$\displaystyle 2 = \frac{2}{3} + C_2$

$\displaystyle \frac{4}{3} = C_2$.

Thus $\displaystyle y(x) = \frac{2}{3}e^{-2x} + \frac{4}{3}xe^{-2x}$.

- Sep 25th 2009, 09:50 AM #3

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- Sep 25th 2009, 05:58 PM #4