What is the solution for y^{'''}-4y^{''}+4y^'=8x ?

The solution for y^{'''}-4y^{''}+4y^'=8x is y=c_{1}+c_{2}e^{2x}+c_{3}xe^{2x}+x^2+2x

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y^{'''}-4y^{''}+4y^'=8x

y^{^{'''}} - 4 y^{^{''}} + 4 y^{^{'}} = 8 x

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

Solution steps

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

Solve linear ODE:

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

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

\int 5 x^{2} e^{- x} d x

\frac{d}{d x} (sin (5 π x))

\frac{d}{d x} (12 ln (\frac{x}{20}))

\int 1 - x^{3} d x

f (x) = (x + 2 x) (2 x + 1)

What is the solution for y^{'''}-4y^{''}+4y^'=8x ?
The solution for y^{'''}-4y^{''}+4y^'=8x is y=c_{1}+c_{2}e^{2x}+c_{3}xe^{2x}+x^2+2x