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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{3 t}, y (0) = 3, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} = 5 x

(D^{2} + 2 D - 3) y = cos (2 x)

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