Questions: Find the solution of the differential equation [ y''-3y'+2y=(3-4x)e^3x ] satisfying the initial conditions (y(0)=frac32) and (y'(0)=frac92). Show your working.

Solution Steps

To solve the given differential equation, we will use the method of undetermined coefficients. First, solve the homogeneous equation \( y'' - 3y' + 2y = 0 \) to find the complementary solution. Then, find a particular solution to the non-homogeneous equation by assuming a form based on the right-hand side \((3-4x)e^{3x}\). Finally, apply the initial conditions to determine the constants in the general solution.

We start by solving the homogeneous part of the differential equation:

\[ y'' - 3y' + 2y = 0 \]

The characteristic equation is:

\[ r^2 - 3r + 2 = 0 \]

Factoring gives:

\[ (r - 1)(r - 2) = 0 \]

Thus, the roots are \( r_1 = 1 \) and \( r_2 = 2 \). The complementary solution is:

\[ y_c = C_1 e^{x} + C_2 e^{2x} \]

Next, we find a particular solution \( y_p \) for the non-homogeneous equation:

\[ y'' - 3y' + 2y = (3 - 4x)e^{3x} \]

We assume a particular solution of the form:

\[ y_p = (Ax + B)e^{3x} \]

Calculating the derivatives and substituting into the differential equation allows us to solve for the coefficients \( A \) and \( B \).

The general solution is given by:

\[ y = y_c + y_p = C_1 e^{x} + C_2 e^{2x} + (Ax + B)e^{3x} \]

Using the initial conditions \( y(0) = \frac{3}{2} \) and \( y'(0) = \frac{9}{2} \), we can solve for \( C_1 \) and \( C_2 \). The final solution, after applying the initial conditions, is:

\[ y = \left( \frac{9}{2} - 2x \right)e^{2x} - 4.0 e^{x} + 1.0 e^{x} \]

Final Answer