What is the solution for y^{''}-2y^'+y=e^tln(t) ?

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

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

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

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{2 e ^{t} t ^{2} ln ( t ) - 3 e ^{t} t ^{2}}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{2 e ^{t} t ^{2} ln ( t ) - 3 e ^{t} t ^{2}}{4}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{2 e ^{t} t ^{2} ln ( t ) - 3 e ^{t} t ^{2}}{4}

y^{^{''}} + 16 y^{^{'}} + 64 y = t^{- 2} e^{- 8 t}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{3 x} + 2

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

y^{^{''}} - 2 y^{^{'}} - 15 = 0

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