What is the solution for y^{''}-2y^'+y=e^t+t,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-2y^'+y=e^t+t,y(0)=1,y^'(0)=0 is y=(e^tt^2)/2+t-e^t+2

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y^{''}-2y^'+y=e^t+t,y(0)=1,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + y = e^{t} + t, y (0) = 1, y^{^{'}} (0) = 0

y = \frac{e ^{t} t ^{2}}{2} + t - e^{t} + 2

Solution steps

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

Solve linear ODE:

y = \frac{e ^{t} t ^{2}}{2} + t - e^{t} + 2

y = \frac{e ^{t} t ^{2}}{2} + t - e^{t} + 2

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

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

y^{^{''}} + y^{^{'}} - 6 y = 4 t e^{2 t}

y^{^{''}} - 9 y = e^{3 x}, y (0) = 3, y^{^{'}} (0) = 10

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} + cos (2 x)

What is the solution for y^{''}-2y^'+y=e^t+t,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-2y^'+y=e^t+t,y(0)=1,y^'(0)=0 is y=(e^tt^2)/2+t-e^t+2