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

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Homework Help Overview

The discussion revolves around finding a fundamental set of solutions for a second-order differential equation represented as 64y'' + 144y' = 0, with specified initial conditions for two functions, y1 and y2. The participants are exploring the characteristic equation and the general solution format.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of the characteristic equation and the general solution. Questions arise regarding the correctness of their expressions for y1 and its derivative, y'1. There is also exploration of the implications of initial conditions on the constants c1 and c2.

Discussion Status

Some participants express uncertainty about their algebraic manipulations and seek clarification on their approach. Guidance is offered regarding potential arithmetic errors, but no consensus on the correct values for c1 and c2 has been reached.

Contextual Notes

Participants note that the problem involves specific initial conditions and that there may be mistakes in the algebraic steps taken to solve for the constants. The discussion reflects a collaborative effort to identify and rectify these errors without providing direct solutions.

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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