Find y_p and the specific solution for y"-3y'-4y=3e^(2t) with y(0)=0 and y'(0)=0

Question

Find y_p and the specific solution for y"-3y'-4y=3e^(2t) with y(0)=0 and y'(0)=0

Answer 1

Characteristic equation r² - 3r - 4 = 0; r = -1 and r = 4. So, general solution of homogenous equation:

y_H = C₁e^-t + C₂e^4t_.

Particular solution of given equation y_P = Ae^2t; y'_p = 2Ae^2t; y''_P = 4Ae^2t;

4Ae^2t - 6Ae^2t - 4Ae^2t = 3e^2t; - 6A = 3; A = - 1/2. So y_P = - 1/2e^2t.

y = C₁e^{- t} + C₂e^4t.- 1/2e^2t. y(0) = C₁ + C₂ - 1/2 = 0; y' = - C₁e^{- t} + 4C₂e^4t - e^2t;

y'(0) = - C₁ + 4C₂ - 1 = 0; 5C₂ = 3/2; C₂ = 3/10, C₁ = 1/5

Solution: y = 1/5e^{- t} + 3/10e^4t - 1/2 e^2t