What is the solution for y^{''}-5y^'+6y=xe^{3x} ?

The solution for y^{''}-5y^'+6y=xe^{3x} is y=c_{1}e^{3x}+c_{2}e^{2x}+(e^{3x}x^2)/2-e^{3x}x

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y^{''}-5y^'+6y=xe^{3x}

y^{^{''}} - 5 y^{^{'}} + 6 y = x e^{3 x}

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = x e^{3 x}

Solve linear ODE:

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

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

y^{^{''}} + 4 y = - cos (3 x) - 2 sin (3 x)

y^{^{''}} + 4 y^{^{'}} = t^{2} + 6 e^{t}

y^{^{''}} - 9 y^{^{'}} = 5 e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 4 y = 4 x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (3 x) cos (2 x)

What is the solution for y^{''}-5y^'+6y=xe^{3x} ?
The solution for y^{''}-5y^'+6y=xe^{3x} is y=c_{1}e^{3x}+c_{2}e^{2x}+(e^{3x}x^2)/2-e^{3x}x