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

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

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^{^{'''}} - 81 y^{^{'}} = t e^{- 9 t} + 10 cos (9 t)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y - e^{x} = 0

y^{^{''}} + 16 y = 5 sec (4 t)

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

x^{^{''}} + x^{^{'}} + x = 1, x^{^{'}} (0) = 0

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