What is the solution for y^{''}-2y^'+y=13e^x ?

The solution for y^{''}-2y^'+y=13e^x is y=c_{1}e^x+c_{2}xe^x+(13e^xx^2)/2

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{13 e ^{x} x ^{2}}{2}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 13 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{13 e ^{x} x ^{2}}{2}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{13 e ^{x} x ^{2}}{2}

y^{^{''}} - y^{^{'}} = e^{x} + 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 t

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 t e^{2 t}, y^{^{'}} (0) = - 3

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

x^{^{''}} + 2 x^{^{'}} - 8 x = 12 e^{2 t} + 8 t - 10

What is the solution for y^{''}-2y^'+y=13e^x ?
The solution for y^{''}-2y^'+y=13e^x is y=c_{1}e^x+c_{2}xe^x+(13e^xx^2)/2