What is the solution for y^{''}-6y^'+3y=xe^x ?

The solution for y^{''}-6y^'+3y=xe^x is y=c_{1}e^{(3+sqrt(6))x}+c_{2}e^{(3-sqrt(6))x}-(e^xx)/2+e^x

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y^{''}-6y^'+3y=xe^x

y^{^{''}} - 6 y^{^{'}} + 3 y = x e^{x}

y = c_{1} e^{(3 + 6) x} + c_{2} e^{(3 - 6) x} - \frac{e ^{x} x}{2} + e^{x}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 3 y = x e^{x}

Solve linear ODE:

y = c_{1} e^{(3 + 6) x} + c_{2} e^{(3 - 6) x} - \frac{e ^{x} x}{2} + e^{x}

y = c_{1} e^{(3 + 6) x} + c_{2} e^{(3 - 6) x} - \frac{e ^{x} x}{2} + e^{x}

3 y^{^{''}} + 8 y^{^{'}} + 4 y = 8 x^{2} - 8 x + 2 + e^{- 2 x}

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (3 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{t} sec (t)

y^{^{''}} + 2 y^{^{'}} + y = t, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 3 y^{^{'}} - 10 y = 6 cos (2 t)

What is the solution for y^{''}-6y^'+3y=xe^x ?
The solution for y^{''}-6y^'+3y=xe^x is y=c_{1}e^{(3+sqrt(6))x}+c_{2}e^{(3-sqrt(6))x}-(e^xx)/2+e^x