# Homework Help: Check DE general solution with complex roots

1. Aug 13, 2010

### boneill3

1. The problem statement, all variables and given/known data

Solve $y''+4y'+5y=0$

find solutions for y(0)=1 and y'(0)=0

2. Relevant equations

3. The attempt at a solution

Hows this look ?

assume solution is in the form of $y=ce^{rx}$

substitute $y=ce^{rx}$ into the equation.

$cr^2e^{rx}+4cre^{rx}+5ce^{rx}=0$

we then divide by $ce^{rx}$ to give

$r^2+4r+5=0$

So we than find the roots which are $r_1=-2+i,r_2=-2-i$

which gives the solution of

$y=c_1e^{-2+ix}+c_2e^{-2-ix}$.

but as they are complex roots we use the equation

$y=c_1e^{\lambda x}cos \mu x +c_2e^{\lambda x}sin \mu x }$.

where $\mu = 1 , \lambda = -2$

so

$y(x)=c_1e^{-2x}cos x +c_2e^{-2x}sin x$.

and

$y'(x)=e^{-2x}(sin (x)(-c_1-sc_2)+cos(x)(c_2-2c_1))$.

to find $c_1$ and $c_2$ apply the initial conditions.

$y(0) = c_1 = 1$
and
$y'(0) = c_2-2= 0$ so $c_2 = 2$

so that makes the general solution

$y(x)=e^{-2x}cos x +2e^{-2x}sin x$.

Last edited: Aug 13, 2010
2. Aug 13, 2010

### Staff: Mentor

It's easy to check.
1) Verify that y(0) = 1 and y'(0) = 2
2) Verify that your solution satisfies y'' + 4y' + 5y = 0.

3. Aug 15, 2010

### beetle2

It's easy to check.
2) Verify that your solution satisfies y'' + 4y' + 5y = 0.

I got
y'' = -5
y' = 0
y =1

so its the solution thanks alot