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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 e^{6 x}

\frac{d ^{2} x}{d t ^{2}} + x k + a = 0

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

y^{^{''}} + 2 y^{^{'}} + y = x^{2} - 5

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

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