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

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

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

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

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^{^{''}} - y = 2 e^{x} - 1 - 6 \cdot x \cdot e^{x}

y^{^{'''}} + 4 y^{^{''}} - 5 y = x e^{x} + 17 x

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 cos (2 t), x^{0} = 1, x^{^{'}} = - 1

y^{^{''}} + \frac{32}{3} y = cos (x)

y^{^{''}} + 4 y = t, y (0) = 0, y^{^{'}} (0) = 0

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