What is the solution for y^{''}-4y^'+3y=8e^{-x}+9 ?

The solution for y^{''}-4y^'+3y=8e^{-x}+9 is y=c_{1}e^{3x}+c_{2}e^x+e^{-x}+3

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y^{''}-4y^'+3y=8e^{-x}+9

y^{^{''}} - 4 y^{^{'}} + 3 y = 8 e^{- x} + 9

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 8 e^{- x} + 9

Solve linear ODE:

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

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

y^{^{''}} + 18 y^{^{'}} + 82 y = 2 e^{- 9 x} cos (x)

y^{^{''}} + 9 y^{^{'}} = t^{2} e^{3 t} + 6

\frac{d ^{2} y}{d t ^{2}} - 12 \frac{d y}{d t} + 36 y = 7 + 4 t

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{(2 x)} \cdot sin (x)

What is the solution for y^{''}-4y^'+3y=8e^{-x}+9 ?
The solution for y^{''}-4y^'+3y=8e^{-x}+9 is y=c_{1}e^{3x}+c_{2}e^x+e^{-x}+3