Questions: Solve the IVP: y’’ + y = 5e^(-4x), y(0) = 0, y’(0) = 0

Solution Steps

To solve the given initial value problem (IVP), we need to find the particular solution to the non-homogeneous differential equation \( y'' + y = 5e^{-4x} \) and then apply the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \). The solution involves finding the complementary solution to the homogeneous equation \( y'' + y = 0 \) and a particular solution to the non-homogeneous equation. The general solution is the sum of these two solutions. Finally, we apply the initial conditions to determine the constants.

We start with the initial value problem (IVP) given by the differential equation:

\[ y'' + y = 5e^{-4x} \]

with initial conditions:

\[ y(0) = 0, \quad y'(0) = 0 \]

The complementary solution is found by solving the homogeneous equation:

\[ y'' + y = 0 \]

The characteristic equation is:

\[ r^2 + 1 = 0 \]

which has roots \( r = i \) and \( r = -i \). Thus, the complementary solution is:

\[ y_c = C_1 \cos(x) + C_2 \sin(x) \]

To find a particular solution \( y_p \) for the non-homogeneous part \( 5e^{-4x} \), we can use the method of undetermined coefficients. We assume a solution of the form:

\[ y_p = Ae^{-4x} \]

Substituting \( y_p \) into the differential equation and solving for \( A \), we find:

\[ y_p = \frac{5}{17} e^{-4x} \]

The general solution is the sum of the complementary and particular solutions:

\[ y = y_c + y_p = C_1 \cos(x) + C_2 \sin(x) + \frac{5}{17} e^{-4x} \]

Using the initial conditions to solve for \( C_1 \) and \( C_2 \):

1. From \( y(0) = 0 \):

\[ C_1 + \frac{5}{17} = 0 \implies C_1 = -\frac{5}{17} \]

2. From \( y'(0) = 0 \):

Calculating \( y' \):

\[ y' = -C_1 \sin(x) + C_2 \cos(x) - \frac{20}{17} e^{-4x} \]

Setting \( y'(0) = 0 \):

\[ C_2 - \frac{20}{17} = 0 \implies C_2 = \frac{20}{17} \]

Final Answer