What is the solution for y^{''''}-y^{''}=4x+2xe-x ?

The solution for y^{''''}-y^{''}=4x+2xe-x is y=c_{1}e^x+c_{2}e^{-x}+c_{3}+c_{4}x-((2e+3)x^3)/6

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y^{''''}-y^{''}=4x+2xe-x

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

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} + c_{4} x - \frac{( 2 e + 3 ) x ^{3}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} + c_{4} x - \frac{( 2 e + 3 ) x ^{3}}{6}

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} + c_{4} x - \frac{( 2 e + 3 ) x ^{3}}{6}

y^{^{''}} + 4 y^{^{'}} - 2 y = \frac{1}{x ^{2} e ^{2 x}}

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

y^{^{''}} - y^{^{'}} + y = 4, y^{^{'}} (0) = - 1, y (0) = 4

y^{^{''}} + y - 2 y = 2 + 5 cos (x)

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

What is the solution for y^{''''}-y^{''}=4x+2xe-x ?
The solution for y^{''''}-y^{''}=4x+2xe-x is y=c_{1}e^x+c_{2}e^{-x}+c_{3}+c_{4}x-((2e+3)x^3)/6