What is the solution for y^{''}-4y=8e^{2x},y(0)=1,y(0)=0 ?

The solution for y^{''}-4y=8e^{2x},y(0)=1,y(0)=0 is y=(1-c_{2})e^{2x}+c_{2}e^{-2x}+2xe^{2x}

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y^{''}-4y=8e^{2x},y(0)=1,y(0)=0

y^{^{''}} - 4 y = 8 e^{2 x}, y (0) = 1, y (0) = 0

y = (1 - c_{2}) e^{2 x} + c_{2} e^{- 2 x} + 2 x e^{2 x}

Solution steps

y^{''} (x) - 4 y = 8 e^{2 x}

Solve linear ODE:

y = (1 - c_{2}) e^{2 x} + c_{2} e^{- 2 x} + 2 x e^{2 x}

y = (1 - c_{2}) e^{2 x} + c_{2} e^{- 2 x} + 2 x e^{2 x}

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

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

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

\frac{d ^{2} y}{d x ^{2}} + 10 \frac{d y}{d x} + 16 y = 40 cos (x)

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

What is the solution for y^{''}-4y=8e^{2x},y(0)=1,y(0)=0 ?
The solution for y^{''}-4y=8e^{2x},y(0)=1,y(0)=0 is y=(1-c_{2})e^{2x}+c_{2}e^{-2x}+2xe^{2x}