What is the solution for y^{''}-16y=32+45e^x,y(0)=0,y^{''}(0)=1 ?

The solution for y^{''}-16y=32+45e^x,y(0)=0,y^{''}(0)=1 is y=(-c_{2}+5)e^{4x}+c_{2}e^{-4x}-2-3e^x

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y^{''}-16y=32+45e^x,y(0)=0,y^{''}(0)=1

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{''}} (0) = 1

y = (- c_{2} + 5) e^{4 x} + c_{2} e^{- 4 x} - 2 - 3 e^{x}

Solution steps

y^{''} (x) - 16 y = 32 + 45 e^{x}

Solve linear ODE:

y = (- c_{2} + 5) e^{4 x} + c_{2} e^{- 4 x} - 2 - 3 e^{x}

y = (- c_{2} + 5) e^{4 x} + c_{2} e^{- 4 x} - 2 - 3 e^{x}

y^{^{''}} - 4 y^{^{'}} - 32 y = - 80 t + 64 t^{2}

D^{3} y = e^{3 x}

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

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

- y^{^{''}} + y = e^{x}

What is the solution for y^{''}-16y=32+45e^x,y(0)=0,y^{''}(0)=1 ?
The solution for y^{''}-16y=32+45e^x,y(0)=0,y^{''}(0)=1 is y=(-c_{2}+5)e^{4x}+c_{2}e^{-4x}-2-3e^x