What is the solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=6 ?

The solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=6 is y=e^x-e^{-3x}+e^{2x}

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y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=6

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 6

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 5 t^{2}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 7

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 6

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 5

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = \frac{e ^{t}}{t}

What is the solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=6 ?
The solution for y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=6 is y=e^x-e^{-3x}+e^{2x}