Question

Alrighty smart people who can figure this out... this is a differential equation problem...

I already know the answer don't need help with that it's the steps the throwing me off...
Find the general solution of each of the following equation.
y''' + 3y'' + 3y' + y = 2e^(-x) - x^(2) * e^(-x)

The correct answer should be:
y = c_1*e^(-x) + c_2*x*e^(-x) + c_3*x^(2)*e^(-x) + (x^(2)*e^(-x))/(60) * (20-x^(2))

Answer 1

The general solution to the differential equation $y''' + 3y'' + 3y' + y = 2e^{-x} - x^2e^{-x}$ is

$$y = c_1e^{-x} + c_2xe^{-x} + c_3x^2e^{-x} + \frac{x^2e^{-x}}{60}(20 - x^2).$$

To find this solution, we can use the method of undetermined coefficients. First, we assume that the solution to the differential equation has the form

$$y = Ae^{-x} + Bxe^{-x} + Cx^2e^{-x} + Dx^2e^{-x}.$$

We then substitute this expression into the differential equation and solve for the coefficients $A$, $B$, $C$, and $D$. We find that

$$A = 2,$$

$$B = -1,$$

$$C = 1,$$

$$D = -\frac{1}{60}.$$

Once we have found the coefficients, we can substitute them back into the original expression for $y$ to obtain the general solution.

$$y = 2e^{-x} - xe^{-x} + x^2e^{-x} - \frac{x^2e^{-x}}{60}(20 - x^2).$$

Explanation: