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^{^{''}} + 4 y = x cos (x)

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

y^{^{''}} + y^{^{'}} - 6 y = 6 x^{3} - 3 x^{2} + 12 x

y^{^{''}} + 6 y^{^{'}} + 12 y = 763 e^{2 t} cos (7 t)

y^{^{''}} + y = \frac{2}{cos ( x )}, y (0) = 2, y^{^{'}} (0) = 2

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