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

In summary, the equation y1=c1*e^(-9/4 *t) + c2*e^(r2*t) has y1=c1e^(-9/4 *t) + c2(1). However, when solving for y'1, you get y'1=(-9/4)c1e^(-9/4 *t) + 0(c2).
  • #1
tristyn
2
0

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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  • #2
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.
 
  • #3
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...
 
  • #4
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.
 

What is a fundamental set of solutions for a 2nd order differential equation?

A fundamental set of solutions for a 2nd order differential equation is a set of linearly independent solutions that can be used to represent all other solutions to the equation. It is typically composed of two distinct solutions.

Why is it important to find a fundamental set of solutions for a 2nd order differential equation?

Finding a fundamental set of solutions allows us to fully understand and solve the given differential equation. It provides a complete set of solutions that can be used to solve for the general solution, as well as any specific solutions that may be needed.

How do you find a fundamental set of solutions for a 2nd order differential equation?

To find a fundamental set of solutions, one can use the method of undetermined coefficients or variation of parameters. Both methods involve finding two distinct solutions to the given equation and then combining them to create a fundamental set.

What is the difference between a general solution and a fundamental set of solutions?

A general solution is an equation that contains all possible solutions to a given differential equation. A fundamental set of solutions, on the other hand, is a set of two distinct solutions that can be used to represent all other solutions to the equation. In other words, the fundamental set of solutions is a subset of the general solution.

Can a fundamental set of solutions be used for any 2nd order differential equation?

Yes, a fundamental set of solutions can be used for any 2nd order differential equation. However, the method used to find the solutions may vary depending on the specific equation. It is important to choose a method that is appropriate for the given equation.

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