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

The solution for y^{''}-3y^'+y=-e^x is y=c_{1}e^{((3+sqrt(5))x)/2}+c_{2}e^{((3-sqrt(5))x)/2}+e^x

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

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

y = c_{1} e^{\frac{( 3 + 5 ) x}{2}} + c_{2} e^{\frac{( 3 - 5 ) x}{2}} + e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 3 + 5 ) x}{2}} + c_{2} e^{\frac{( 3 - 5 ) x}{2}} + e^{x}

y = c_{1} e^{\frac{( 3 + 5 ) x}{2}} + c_{2} e^{\frac{( 3 - 5 ) x}{2}} + e^{x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{1 - x ^{2}}

y^{^{''}} - 5 y^{^{'}} + 4 y = x, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 35 y = (t + 1) e^{t}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 4 e^{3 x} + 8 x

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

What is the solution for y^{''}-3y^'+y=-e^x ?
The solution for y^{''}-3y^'+y=-e^x is y=c_{1}e^{((3+sqrt(5))x)/2}+c_{2}e^{((3-sqrt(5))x)/2}+e^x