Solve the differential equation using the method of undetermined coefficients.

Question

Quizplus · Accepted Answer

The differential equation is of the form:

$$
y'' - 2y' + y = 2e^x
$$

The characteristic equation is:

$$
r^2 - 2r + 1 = 0 
$$

Solving this, we get:

$$
r = 1
$$

Now, for the particular solution, since $2e^x$ is already a solution of the homogeneous equation, we will try the method of undetermined coefficients using $xe^x$ as the guess for the particular solution.

Thus, we have:

$$
y_p = Ax e^x 
$$

Taking the first and second derivatives, we have:

$$
y_p' = (A + Ax) e^x 
$$

and

$$
y_p'' = (2A + Ax) e^x 
$$

Plugging these into the original differential equation, we get:

$$
y_p'' - 2y_p' + y_p = 2e^x
$$

$$(2A + Ax) e^x - 2(A + Ax)e^x + Axe^x = 2e^x$$

Simplifying, we have:

$$
A = 2
$$

So, the particular solution is:

$$
y_p = 2xe^x
$$

Therefore, the general solution is:

$$
y = c_1 e^x + c_2 xe^x + 2xe^x
$$

which is choice (B).

Solve the Differential Equation Using the Method of Undetermined Coefficients