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}

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