What is the solution for (d^2y)/(dx^2)-4y^'+3y=8e^{-x}+9 ?

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

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(d^2y)/(dx^2)-4y^'+3y=8e^{-x}+9

\frac{d ^{2} y}{d x ^{2}} - 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

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