# Finding a fundamental set of solutions for a 2nd order differential equation

1. Oct 12, 2012

### tristyn

1. The problem statement, all variables and given/known data
64y''+144y'=0

y1(0)=1 y'1(0)=0
and
y2(0)=0 and y'2(0)=1

2. Relevant equations
y1=c1*e^(r1*t) + c2*e^(r2*t)

3. The attempt at a solution

I start by finding the characteristic equation:
64r^2+144r=0
r1=-9/4 and r2=0

y1=c1e(r1*t) + c2e(r2*t)

so I get
y1=c1e^(-9/4 *t) + c2e^(0*t)

e^(0*t) = 1 will always = 0, which gives
y1=c1e^(-9/4 *t) + c2(1)

so I suppose I am asking if I started this wrong or if not, because I need y'1.

With these values I would have:
y'1=(-9/4)c1e^(-9/4 *t) + 0(c2)
?? because the derivative of 1 is zero

Is this correct or have I gone about the problem in the wrong way?

2. Oct 12, 2012

### fluidistic

This looks like correct to me.

3. Oct 12, 2012

### tristyn

Okay, so when I go from there
y=c1*e^(-9/4 *t) + c2
y'=(-9/4)*e^(-9/4 *t)*c1 + c2??

1=c1+c2
0=(-9/4)c1+c2
for y(0)=1 y'(0)=0

c1=(1-c2)
0=(-9/4)(1-c2) + c2
9/4=(13/4)*c2
c2=9/13
c1=4/13

but this isn't correct according to webwork, so what is the mistake? is it supposed to be

y=c1*e^(-9/4 *t) + c2
y'=(-9/4)*e^(-9/4 *t)*c1
giving
1=c1+c2
0=(-9/4)c1
c1=(1-c2)
0=(-9/4)(1-c2)
c2=9/4
c1=-5/4

but that isn't right either?

I'm just stumped on this, and it is probably a rather silly mistake somewhere in my algebra or something...

4. Oct 12, 2012

### fluidistic

As you noted the derivative of a constant is 0 so this is wrong.
You might want to redo the arithmetics in the red part. :)
Other than this, you're doing fine.