What is the solution for y^{''}-2y^'+y=x^3+4x ?

The solution for y^{''}-2y^'+y=x^3+4x is y=c_{1}e^x+c_{2}xe^x+x^3+6x^2+22x+32

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 22 x + 32

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 22 x + 32

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 22 x + 32

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

- \frac{1}{9} y^{^{''}} + \frac{2}{3} y^{^{'}} - y = \frac{t}{t + 1}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = t e^{- 3 t}

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

What is the solution for y^{''}-2y^'+y=x^3+4x ?
The solution for y^{''}-2y^'+y=x^3+4x is y=c_{1}e^x+c_{2}xe^x+x^3+6x^2+22x+32