What is the solution for y^{''}-4y=3e^{2x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y=3e^{2x},y(0)=0,y^'(0)=0 is y=-3/16 e^{2x}+3/16 e^{-2x}+(3e^{2x}x)/4

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y^{''}-4y=3e^{2x},y(0)=0,y^'(0)=0

y^{^{''}} - 4 y = 3 e^{2 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{3}{16} e^{2 x} + \frac{3}{16} e^{- 2 x} + \frac{3 e ^{2 x} x}{4}

Solution steps

y^{''} (x) - 4 y = 3 e^{2 x}

Solve linear ODE:

y = - \frac{3}{16} e^{2 x} + \frac{3}{16} e^{- 2 x} + \frac{3 e ^{2 x} x}{4}

y = - \frac{3}{16} e^{2 x} + \frac{3}{16} e^{- 2 x} + \frac{3 e ^{2 x} x}{4}

2 y^{^{''}} + 3 y^{^{'}} - 2 y = t e^{- 2 t}

x^{^{''}} + 0.8 x^{^{'}} + 19.6 x = 4 e^{- 2 t} cos (6 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 0, y^{^{'}} (0) = 1

(d^{2} + 2 d + 3) y = e^{2} x sin (x) + 3

(d - 2)^{3} y = e^{2} x

What is the solution for y^{''}-4y=3e^{2x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y=3e^{2x},y(0)=0,y^'(0)=0 is y=-3/16 e^{2x}+3/16 e^{-2x}+(3e^{2x}x)/4