What is the solution for y^{''}+2y^'+10=0,y(0)=-2,y^'(0)=0 ?

The solution for y^{''}+2y^'+10=0,y(0)=-2,y^'(0)=0 is y=-2+5/2-5/2 e^{-2t}-5t

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y^{''}+2y^'+10=0,y(0)=-2,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 10 = 0, y (0) = - 2, y^{^{'}} (0) = 0

y = - 2 + \frac{5}{2} - \frac{5}{2} e^{- 2 t} - 5 t

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 = 0

Solve linear ODE:

y = - 2 + \frac{5}{2} - \frac{5}{2} e^{- 2 t} - 5 t

y = - 2 + \frac{5}{2} - \frac{5}{2} e^{- 2 t} - 5 t

4 y^{^{''}} + y = 10 cos (x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 e^{- 2} x

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 x + 3

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (x) + 4 x^{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = 3 x^{2} - 4 x

What is the solution for y^{''}+2y^'+10=0,y(0)=-2,y^'(0)=0 ?
The solution for y^{''}+2y^'+10=0,y(0)=-2,y^'(0)=0 is y=-2+5/2-5/2 e^{-2t}-5t