Easy DEQ problem(s)

1. Jan 9, 2012

QuarkCharmer

1. The problem statement, all variables and given/known data
Verify that the indicated function is a solution to the differential equation. Assume an appropriate interval I of definition for each solution.

$$y'' - 6y' +13y = 0$$
$$y = e^{3x}cos(2x)$$

2. Relevant equations

3. The attempt at a solution
Well, I started taking derivatives to see if y was indeed a solution, but that got rather complicated fast for what is some of the first problem sets from day 1 of class. I believe that this is NOT a solution because of the cosine, which would indicate that y(x) is not a linear function clearly. Since the DEQ is linear of the second order, I assume that the solution will be linear as well?

Frankly, I am confused on when to consider a function linear. Does the dependent variable have to be linear, or is it the independent variable that matters? What if I replaced cos(2x) in the above equation with cos(2y).

Is this the correct manner of thinking on this problem? (ie: see cosine, must not be correct)

2. Jan 9, 2012

HACR

it is the solution to the homogeneous diff equ. Linear means that none of the terms are of some power like squared.

3. Jan 9, 2012

vela

Staff Emeritus
I think you need to recalibrate your sense of what qualifies as complicated.

No, that's not correct.

You're not interested in whether a function is linear here. You want to know if an operator is linear. The equation you have can be written in the form $\mathcal{L}y = f(x)$, where $\mathcal{L}$ is an operator acting on y. If the operator is linear, it has the properties
\begin{align*}
\mathcal{L}(y_1 + y_2) &= \mathcal{L}y_1 + \mathcal{L}y_2 \\
\mathcal{L}(cy) &= c\mathcal{L}y
\end{align*}and the concepts from linear algebra can be applied to solve the equation.

To determine if the operator $\mathcal{L}$ is linear, you just need to make sure that y and its derivatives, if they appear, are only taken to the first power, which is indeed the case here.