\( \left\{ \begin{array} { l } { y ^ { \prime \prime } + 3 y ^ { \prime } - 4 y = 5 e ^ { x } } \\ { y ( 0 ) = 2 , y ^ { \prime } ( 0 ) = 1 } \end{array} \right. \)

To solve the differential equation \[ y'' + 3y' - 4y = 5e^x \] with the initial conditions \( y(0) = 2 \) and \( y'(0) = 1 \), we will use the method of undetermined coefficients. ### Step 1: Solve the Homogeneous Equation First, we solve the associated homogeneous equation: \[ y'' + 3y' - 4y = 0 \] The characteristic equation is: \[ r^2 + 3r - 4 = 0 \] To find the roots, we can use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \] Calculating the roots: \[ r_1 = 1, \quad r_2 = -4 \] Thus, the general solution to the homogeneous equation is: \[ y_h(x) = C_1 e^{x} + C_2 e^{-4x} \] ### Step 2: Find a Particular Solution Next, we find a particular solution \( y_p \) to the non-homogeneous equation. Since the right-hand side is \( 5e^x \), we will try a particular solution of the form: \[ y_p = A x e^x \] We choose \( A x e^x \) because \( e^x \) is a solution to the homogeneous equation, and we need to multiply by \( x \) to find a suitable form. Now we compute the derivatives: \[ y_p' = A(e^x + xe^x) = A e^x (1 + x) \] \[ y_p'' = A(e^x + e^x + xe^x) = A e^x (2 + x) \] Substituting \( y_p, y_p', y_p'' \) into the original equation: \[ A e^x (2 + x) + 3A e^x (1 + x) - 4A x e^x = 5 e^x \] Simplifying this: \[ A e^x (2 + x + 3 + 3x - 4x) = 5 e^x \] \[ A e^x (5) = 5 e^x \] Thus, we have: \[ 5A = 5 \implies A = 1 \] So, the particular solution is: \[ y_p = x e^x \] ### Step 3: General Solution The general solution to the differential equation is: \[ y(x) = y_h(x) + y_p(x) = C_1 e^x + C_2 e^{-4x} + x e^x \] ### Step 4: Apply Initial Conditions Now we apply the initial conditions to find \( C_1 \) and \( C_2 \). 1. From \( y(0) = 2 \): \[ C_1 e^0 + C_2 e^0 + 0 \cdot e^0 = 2 \implies C_1 + C_2 = 2 \] 2. From \( y'(0) = 1 \): First, we compute \( y' \): \[ y' = C_1 e^x - 4C_2 e^{-4x} + e^x + x e^x \] Evaluating at \( x = 0 \): \[ y'(0) = C_1 - 4C_2 + 1 + 0 = 1 \implies C_1 - 4C_2 + 1 = 1 \implies C_1 - 4C_2 = 0 \] ### Step 5: Solve the System of Equations Now we have the system: 1. \( C_1 + C_2 = 2 \) 2. \( C_1 - 4C_2 = 0 \) From the second equation, we can express \( C_1 \): \[ C_1 = 4C_2 \] Substituting into the first equation: \[ 4C_2 + C_2 = 2 \implies 5C_2 = 2 \implies C_2 = \frac{2}{5} \] Then substituting back to find \( C_1 \): \[ C_1 = 4 \cdot \frac{2}{5} = \frac{8}{5} \] ### Final Solution Thus, the solution to the initial value problem is: \[ y(x) = \frac{8}{5} e^x + \frac{2}{5} e^{-4x} + x e^x \]