What is the solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=0 ?

The solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=0 is y=-1/2 e^x+1/2 e^{-3x}+e^{2x}

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y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

y = - \frac{1}{2} e^{x} + \frac{1}{2} e^{- 3 x} + e^{2 x}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 5 e^{2 x}

Solve linear ODE:

y = - \frac{1}{2} e^{x} + \frac{1}{2} e^{- 3 x} + e^{2 x}

y = - \frac{1}{2} e^{x} + \frac{1}{2} e^{- 3 x} + e^{2 x}

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

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 4 y = x - sin (3 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = x cos (2 x)

y^{^{''}} - y^{^{'}} - 64 y = 54 cos (3 x)

What is the solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=0 ?
The solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=0 is y=-1/2 e^x+1/2 e^{-3x}+e^{2x}