Verifying Solution to DEQ: y = e^{3x}cos(2x)

[QuarkCharmer]

Homework Statement

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)

Homework Equations

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)

 

[HACR]

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

 

[vela]

Homework Statement

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)

Homework Equations

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

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?

No, that's not correct.

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).

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.