Check DE general solution with complex roots

[boneill3]

Homework Statement

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

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

Homework Equations

Quadratic equation

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.

 

[Mark44]

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.

 

[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 a lot