What is the solution for y^{''}-2y^'+y=6e^x ?

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

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

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

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 6 e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 13 + e x

y^{^{''}} + 10 y^{^{'}} + 25 y = 30 x + 3

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

y^{^{''}} + 9 y = \frac{3}{2} sin (2 t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 10 e^{3 x}

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