What is the solution for y^{''}-2y^'+y=3e^t ?

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

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

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

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = 3 e^{t}

Solve linear ODE:

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

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

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

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{x}

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

16 y^{^{''}} - 8 y^{^{'}} + y = 33 - 16 x + x^{2}

y^{^{''}} - 8 y^{^{'}} + 7 y = 4 e^{3} x

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