What is the solution for y^{''}-3y^'+2y=xe^{3x} ?

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

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y^{''}-3y^'+2y=xe^{3x}

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{e ^{3 x} x}{2} - \frac{3}{4} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{e ^{3 x} x}{2} - \frac{3}{4} e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{e ^{3 x} x}{2} - \frac{3}{4} e^{3 x}

y^{^{'''}} - 2 y^{^{''}} + 4 y^{^{'}} - 8 y = 8 sin (2 x)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = x ln (x)

y^{^{''}} + 4 y = 16 t, y (0) = 3

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

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

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