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

The solution for y^{''''}-3y^{'''}+2y^{''}=e^{3x} is y=c_{1}e^x+c_{2}e^{2x}+c_{3}+c_{4}x+1/18 e^{3x}

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

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

y = c_{1} e^{x} + c_{2} e^{2 x} + c_{3} + c_{4} x + \frac{1}{18} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{2 x} + c_{3} + c_{4} x + \frac{1}{18} e^{3 x}

y = c_{1} e^{x} + c_{2} e^{2 x} + c_{3} + c_{4} x + \frac{1}{18} e^{3 x}

y^{^{''}} + y^{^{'}} = - cos (x)

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

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

4 y^{^{''}} - 4 y^{^{'}} + y = 40 e^{\frac{t}{2}}

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

What is the solution for y^{''''}-3y^{'''}+2y^{''}=e^{3x} ?
The solution for y^{''''}-3y^{'''}+2y^{''}=e^{3x} is y=c_{1}e^x+c_{2}e^{2x}+c_{3}+c_{4}x+1/18 e^{3x}