Using Characteristic equations to solve 2nd order linear DEQ's

In summary, the problem asks to find the largest interval where the solution to the initial value problem of cos(t/3)y'' +(6t^2)y' + ((t-5)^-3)y = 0, with initial conditions y(1) = 1 and y'(1) = 3, is valid. After converting it to its characteristic equation and attempting to factor it, the individual realizes they may be on the wrong track. However, after considering the nature of the problem, they realize that the solution is simply the interval [1,5) due to the singularity at t=5 and the continuity of the other terms.
  • #1
Tricky557
7
0
The problem statement:

Find the largest interval in which the solution of the following initial value problem is valid:

cos(t/3)y'' +(6t^2)y' + ((t-5)^-3)y = 0

Initial conditions:

y(1) = 1
y'(1)= 3

I have a few questions concerning this problem.

I've converted it to it's characteristic equation:

r^2 + [(6t^2)/(cos(t/3)]r + [(t-5)^-3]/(cos(t/3))

This isn't something I'm readily able to factor. So I move to the quadratic equation... Which is what I'm confused about.

When I write down the quadratic equation for this problem, I have a feeling that I've gone down the wrong path to the solution because it turns out to be something that I don't really know what to do with. I can get 2 different and unique roots, but they're both going to be in terms of t. Every example I've seen of this being done always ends with the 2 r's as numbers. Never in terms of t.

I'm not sure what to do here.

Do I even need to solve this equation? I've been thinking about it and all it asks is for the interval where the solutions will be valid, and I can do that relatively easily. All I would need to do, if that were the case for this problem, is to solve these few equations:

cos(t/3) = 0

t-5 = 0

Wherever these equations are true are where the interval boundaries are going to be.

A long post.. almost a wall of text, any help would be greatly appreciated.
 
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  • #2
Since (t-5)^-3 has a singularity at t=5 & other terms are continuous,the answer is [1,5).
 

1. How do you determine the characteristic equation for a 2nd order linear DEQ?

To determine the characteristic equation, you need to rearrange the given differential equation in the form of a polynomial in terms of the dependent variable and its derivatives. The coefficients of the polynomial will form the characteristic equation.

2. What is the significance of the roots of the characteristic equation?

The roots of the characteristic equation represent the solutions to the given differential equation. These roots can be real or complex, and they determine the type of solution and the behavior of the system.

3. Can the characteristic equation be used to solve any type of 2nd order linear DEQ?

Yes, the characteristic equation can be used to solve any linear 2nd order differential equation with constant coefficients. However, for non-constant coefficients, other methods such as variation of parameters may be required.

4. How do you use the roots of the characteristic equation to find the general solution?

The general solution of a 2nd order linear DEQ can be written as a linear combination of the two linearly independent solutions, which can be found using the roots of the characteristic equation. For real roots, the solutions will be exponential functions, and for complex roots, the solutions will be a combination of sine and cosine functions.

5. Are there any limitations to using characteristic equations to solve 2nd order linear DEQ's?

One limitation is that it can only be used for linear differential equations. Additionally, it may not be applicable for non-homogeneous equations or equations with variable coefficients. In such cases, other methods such as Laplace transforms or numerical methods may be more suitable.

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