What is the solution for y^{''}-2y^'+y=4cos(t) ?

The solution for y^{''}-2y^'+y=4cos(t) is y=c_{1}e^t+c_{2}te^t-2sin(t)

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

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

y = c_{1} e^{t} + c_{2} t e^{t} - 2 sin (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} - 2 sin (t)

y = c_{1} e^{t} + c_{2} t e^{t} - 2 sin (t)

y^{^{''}} + y^{^{'}} - 6 y = - 5 e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- 3 x} + 18 x + 15

y^{^{''}} - 2 y^{^{'}} - 8 y = t e^{- 4 t}

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

2 \frac{d ^{2} x}{d t ^{2}} + 18 x + 5 = 0

What is the solution for y^{''}-2y^'+y=4cos(t) ?
The solution for y^{''}-2y^'+y=4cos(t) is y=c_{1}e^t+c_{2}te^t-2sin(t)