Let's solve the given second-order differential equation step by step: ### **Given:** \[ y'' + 2y' - 3y = 5e^x \] #### **Step 1: Solve the Homogeneous Equation** First, solve the homogeneous part: \[ y'' + 2y' - 3y = 0 \] The characteristic equation is: \[ r^2 + 2r - 3 = 0 \] Factorize: \[ (r + 3)(r - 1) = 0 \] So, the roots are \( r = -3 \) and \( r = 1 \). **Homogeneous Solution:** \[ y_h(x) = C_1 e^{-3x} + C_2 e^{x} \] --- #### **Step 2: Find a Particular Solution (\(y_p\))** Since the right side is \( 5e^x \), try a particular solution of the form: \[ y_p = A e^x \] Compute derivatives: \[ y_p' = A e^x \] \[ y_p'' = A e^x \] Plug into the original equation: \[ A e^x + 2A e^x - 3A e^x = 5e^x \] \[ (A + 2A - 3A) e^x = 5e^x \] \[ 0 \cdot e^x = 5e^x \] Uh-oh! This means \(e^x\) is a solution to the homogeneous equation, so we need to try a new form: Try: \[ y_p = B x e^x \] Compute derivatives: \[ y_p' = B e^x + B x e^x \] \[ y_p'' = B e^x + B e^x + B x e^x = 2B e^x + B x e^x \] Plug into the equation: \[ [2B e^x + B x e^x] + 2[B e^x + B x e^x] - 3[B x e^x] = 5e^x \] Expand: \[ 2B e^x + B x e^x + 2B e^x + 2B x e^x - 3B x e^x = 5e^x \] Combine like terms: - \( e^x \) terms: \(2B + 2B = 4B\) - \( x e^x \) terms: \(B + 2B - 3B = 0\) So: \[ 4B e^x = 5e^x \implies 4B = 5 \implies B = \frac{5}{4} \] So, the particular solution is: \[ y_p = \frac{5}{4} x e^x \] --- #### **Step 3: General Solution** \[ \boxed{ y(x) = C_1 e^{-3x} + C_2 e^{x} + \frac{5}{4} x e^x } \] --- **Summary:** - Homogeneous solution: \( C_1 e^{-3x} + C_2 e^x \) - Particular solution: \( \frac{5}{4} x e^x \) - **General solution:** \(\boxed{y(x) = C_1 e^{-3x} + C_2 e^{x} + \frac{5}{4} x e^x}\)