What is the solution for y^{''}-2y^'-3y=-4e^x+3 ?

The solution for y^{''}-2y^'-3y=-4e^x+3 is y=c_{1}e^{3x}+c_{2}e^{-x}+e^x-1

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = sin (t) + cos (t)

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

y^{^{''}} - y^{^{'}} - 6 y = 12 x - e^{x}

y^{^{''}} + 5 y^{^{'}} - 4 y = 8 t + 8

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

What is the solution for y^{''}-2y^'-3y=-4e^x+3 ?
The solution for y^{''}-2y^'-3y=-4e^x+3 is y=c_{1}e^{3x}+c_{2}e^{-x}+e^x-1