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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{t}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = 9 e^{- x}

y^{^{''}} + 4 y = - 20 sin (2 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{x}

x^{^{''}} + 2 x^{^{'}} + 5 x = 1, x^{^{'}} (0) = 0, x (0) = 0

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