What is the solution for y^{''}-4y^'+3y=e^{3x} ?

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

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

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

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = e^{3 x}

Solve linear ODE:

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

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

y^{^{''}} + 9 y = t^{2} e^{3 t} + 6

y^{^{'''}} + 2 y^{^{''}} - 13 y^{^{'}} + 10 y = x e^{- x}

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

y^{^{''}} + 4 y = 7 sec (2 x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2

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