The solution of the differential equation

\(\frac{{{d^2}y}}{{d{x^2}}} - \frac{{dy}}{{dx}} - 2y = 3{e^{2x}}\)

Where y(0) = 0 and y’(0) = -2 is

(D² – D – 2) y = 3 e^2x

A.E. is. D² – D – 2 = 0

⇒ (D – 2) (D + 1) = 0

⇒ D = -1, 2

C.F. = C₁ e^-x + C₂ e^2x

\(P.I. = \frac{1}{{\left( {D - 2} \right)\left( {D + 1} \right)}}.3{e^{2x}}\)

\( = \frac{1}{{{D^2} - D - 2}}3{e^{2x}}\)

\(f\left( D \right) = {D^2} - D - 2\)

\(f\left( 2 \right) = 0\)

\(f'\left( D \right) = 2D - 1\)

\(f'\left( 2 \right) = 3\)

\( PI= \frac{1}{{{D^2} - D - 2}}3{e^{2x}}=x \frac{1}{{ 2D - 1}}3{e^{2x}}\)

\(P.I.= x.\frac{1}{3}.3{e^{2x}} = x{e^{2x}}\)

Complete solution is

\(y = {C_1}{e^{ - x}} + {C_2}{e^{2x}} + x{e^{2x}}\)

Given that y(0) = 0

⇒ 0 = C₁ + C₂ ⇒ C₁ = -C₂

y’(0) = -2

\(y' = - {C_1}{e^{ - x}} + 2{C_2}{e^{2x}} + {e^{2x}} + 2x{e^{2x}}\)

⇒ -2 = -C₁ + 2C₂ + 1

→ -C₁ + 2C₂ = -3

⇒ -C₁ -2C₁ = - 3 ⇒ C₁ = 1, C₂ = -1