What is the solution for y^{''}-2y^'+y=te^t+4 ?

The solution for y^{''}-2y^'+y=te^t+4 is y=c_{1}e^t+c_{2}te^t+(e^tt^3)/6+4

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y^{''}-2y^'+y=te^t+4

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

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6} + 4

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = t e^{t} + 4

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6} + 4

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6} + 4

y^{^{''}} + 36 y = sec^{2} (6 x)

y^{^{'}} = 3 x^{2} (1 + y^{2})

\frac{d}{d x} (\frac{x - 4}{x} + ln (\frac{x}{4}))

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{1}{3}

\frac{d}{d x} (x^{4} + 8 x)

What is the solution for y^{''}-2y^'+y=te^t+4 ?
The solution for y^{''}-2y^'+y=te^t+4 is y=c_{1}e^t+c_{2}te^t+(e^tt^3)/6+4