# Help :(

• September 25th 2009, 08:46 AM
MissWonder
Help :(
Okay.. Can anyone explain me, how to solve this problem:

$y''+4y'+4=0$.. when $y(0)=2$ and $y'(0)=0$
• September 25th 2009, 08:57 AM
Prove It
Quote:

Originally Posted by MissWonder
Okay.. Can anyone explain me, how to solve this problem:

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

The Characteristic Equation is

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

$(m + 2)^2 = 0$

$m = -2$.

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

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

We can also see that

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

Applying the initial conditions gives

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

Solving the equations simultaneously gives

$2 = 3C_1$

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

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

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

Thus $y(x) = \frac{2}{3}e^{-2x} + \frac{4}{3}xe^{-2x}$.
• September 25th 2009, 09:50 AM
MissWonder
hmm
Applying the initial conditions gives

http://www.mathhelpforum.com/math-he...0452902a-1.gif and http://www.mathhelpforum.com/math-he...8b7791ab-1.gif.

Where do you get that last part from?
• September 25th 2009, 05:58 PM
mr fantastic
Quote:

Originally Posted by MissWonder
Applying the initial conditions gives

http://www.mathhelpforum.com/math-he...0452902a-1.gif and http://www.mathhelpforum.com/math-he...8b7791ab-1.gif.

Where do you get that last part from?

The reply given was quite clear. Go back and read it carefully. Especially the part where the given initial condition y'(0) = 0 is substituted into the rule for y'.