What is the solution for y^{''}-2y^'+y=t*e^t+4,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-2y^'+y=t*e^t+4,y(0)=1,y^'(0)=1 is y=-3e^t+4e^tt+(e^tt^3)/6+4

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y^{''}-2y^'+y=t*e^t+4,y(0)=1,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + y = t \cdot e^{t} + 4, y (0) = 1, y^{^{'}} (0) = 1

y = - 3 e^{t} + 4 e^{t} t + \frac{e ^{t} t ^{3}}{6} + 4

Solution steps

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

Solve linear ODE:

y = - 3 e^{t} + 4 e^{t} t + \frac{e ^{t} t ^{3}}{6} + 4

y = - 3 e^{t} + 4 e^{t} t + \frac{e ^{t} t ^{3}}{6} + 4

y^{^{''}} - y^{^{'}} - 6 y = 33 sin (3 x)

y^{^{'''}} + y^{^{''}} - 2 y = e^{3 x} (9 + 67 x + 17 x^{2})

x^{^{''}} - 3 x^{^{'}} - 28 x = sin (6 t)

y^{^{''}} - 4 y^{^{'}} + 5 y = x e^{2 x}

y^{^{''}} - 8 y^{^{'}} + 80 y = - 2 sin (3 t)

What is the solution for y^{''}-2y^'+y=t*e^t+4,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-2y^'+y=t*e^t+4,y(0)=1,y^'(0)=1 is y=-3e^t+4e^tt+(e^tt^3)/6+4