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

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

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

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

y = 2 e^{t} t + cos (t)

Solution steps

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

Solve linear ODE:

y = 2 e^{t} t + cos (t)

y = 2 e^{t} t + cos (t)

(D^{2} + 1) y = e^{- x} + cos (x) + x^{3}

y^{^{''}} - 3 y^{^{'}} + y = x

y^{^{''}} + y^{^{'}} = x (e^{3 x} + 1)

f^{^{''}} (t) + 9 \cdot f (t) = (- 2 \cdot e^{(3 t)})

y^{^{''}} + 4 y^{^{'}} + 5 y = 10 e^{- 3 x}

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