Homogeneous initial value problem

[mtxop2]

Homework Statement

4y" + 4y' + 5y = 0
y(0) = 3
y'(0) = 1

Homework Equations

yh = e^ax(c1cosbx + c2sinbx)

The Attempt at a Solution

For the roots I got -1/2 + i and -1/2 - i so my a = -1/2 and b = 1

then I have to differentiate yh = e^(-1/2x)[c1cosx + c2sinx]

this is where I get this overly complicated equation and I was wondering if I could do integration by parts instead.

where I could get something like

yh = -1/2e^(-1/2x)[c1cosx + c2sinx] + [-c1sinx + c2cosx]e^(-1/2x)

 

[Mark44]

Your title is misleading. Your equation is a homogeneous equation.

Homework Statement

4y" + 4y' + 5y = 0
y(0) = 3
y'(0) = 1

Homework Equations

yh = e^ax(c1cosbx + c2sinbx)

The Attempt at a Solution

For the roots I got -1/2 + i and -1/2 - i so my a = -1/2 and b = 1

then I have to differentiate yh = e^(-1/2x)[c1cosx + c2sinx]

this is where I get this overly complicated equation and I was wondering if I could do integration by parts instead.

?
You can't use integration by parts if you need to differentiate a function.

where I could get something like

yh = -1/2e^(-1/2x)[c1cosx + c2sinx] + [-c1sinx + c2cosx]e^(-1/2x)

 

[mtxop2]

Your title is misleading. Your equation is a homogeneous equation.
?
You can't use integration by parts if you need to differentiate a function.

Damn I'm so sorry I misled you, I've been working on these all day and well you know how that goes. What I meant to say was product rule, where f(x) = e^(-1/2x) and g(x) = [c1cosx + c2sinx]

 

[ehild]

Yes, use product rule for y= e^-x/2(c₁cosx+c₂sinx) to get y'.

ehild