The given expression for $y$ is:

$y = e^{-x} \left( c_1 \cos 3x + c_2 \sin 3x \right)$

This appears to be the general solution to a second-order linear differential equation with constant coefficients. Specifically, this form suggests the solution to a differential equation where the characteristic equation has complex roots.

Here’s how such a solution is typically derived:

1. Characteristic Equation

For a second-order linear differential equation with constant coefficients of the form:

$y'' + p y' + q y = 0$

The characteristic equation is:

$r^2 + p r + q = 0$

2. Complex Roots

If the roots of the characteristic equation are complex, they will be of the form:

$r = \alpha \pm i\beta$

The general solution to the differential equation is then given by:

$y = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))$

3. Matching with the Given Solution

In your case, the solution is:

$y = e^{-x}(c_1 \cos(3x) + c_2 \sin(3x))$

Here, $\alpha = -1$ and $\beta = 3$, so the corresponding characteristic equation would be:

$r^2 + 2r + 10 = 0$

Thus, the differential equation this solution corresponds to is:

$y'' + 2y' + 10y = 0$

Would you like a more detailed explanation on how to derive this step-by-step?

Related Questions:

1. How do we find the characteristic equation from a given second-order differential equation?
2. What is the role of complex roots in solving differential equations?
3. How does Euler’s formula apply to differential equation solutions with complex roots?
4. What changes in the solution if the roots are real instead of complex?
5. How does the method change if the differential equation is non-homogeneous?

Tip: When dealing with characteristic equations, always check if the roots are real, repeated, or complex to determine the right form of the solution.