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

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The discussion centers on solving the second-order differential equation 64y'' + 144y' = 0 with initial conditions y1(0) = 1, y'1(0) = 0, and y2(0) = 0, y'2(0) = 1. The characteristic equation derived is 64r^2 + 144r = 0, yielding roots r1 = -9/4 and r2 = 0. The solutions for y1 and its derivative y'1 were correctly formulated, but the user encountered difficulties in calculating the constants c1 and c2 based on the initial conditions, leading to confusion about algebraic errors in their calculations.

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tristyn
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Homework Statement


64y''+144y'=0

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

Homework Equations


y1=c1*e^(r1*t) + c2*e^(r2*t)

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?
 
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tristyn said:

Homework Statement


64y''+144y'=0

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

Homework Equations


y1=c1*e^(r1*t) + c2*e^(r2*t)


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?
This looks like correct to me.
 
fluidistic said:
This looks like correct to me.

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

Which leads me to
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...
 
tristyn said:
Okay, so when I go from there
y=c1*e^(-9/4 *t) + c2
y'=(-9/4)*e^(-9/4 *t)*c1 + c2??
As you noted the derivative of a constant is 0 so this is wrong.
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...
You might want to redo the arithmetics in the red part. :)
Other than this, you're doing fine.
 

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