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

The solution for y^{''}-4y=e^{2x},y(0)=0,y^'(0)=0 is y=-1/16 e^{2x}+1/16 e^{-2x}+(e^{2x}x)/4

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

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

y = - \frac{1}{16} e^{2 x} + \frac{1}{16} e^{- 2 x} + \frac{e ^{2 x} x}{4}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{16} e^{2 x} + \frac{1}{16} e^{- 2 x} + \frac{e ^{2 x} x}{4}

y = - \frac{1}{16} e^{2 x} + \frac{1}{16} e^{- 2 x} + \frac{e ^{2 x} x}{4}

y^{^{''}} - 3 y^{^{'}} - 28 y = 1056 e^{- t}

y^{^{''}} + y = g (t), y (0) = 0, y^{^{'}} (0) = 0

x^{^{''}} - x = 4 e^{- t}

y^{^{''}} + 6 y^{^{'}} + 18 y = 11 t e^{4 t}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 8

What is the solution for y^{''}-4y=e^{2x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y=e^{2x},y(0)=0,y^'(0)=0 is y=-1/16 e^{2x}+1/16 e^{-2x}+(e^{2x}x)/4