What is the solution for y^{''}+2y^'+17=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+2y^'+17=0,y(0)=1,y^'(0)=-1 is y=1+15/4-15/4 e^{-2t}-(17t)/2

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y^{''}+2y^'+17=0,y(0)=1,y^'(0)=-1

y^{^{''}} + 2 y^{^{'}} + 17 = 0, y (0) = 1, y^{^{'}} (0) = - 1

y = 1 + \frac{15}{4} - \frac{15}{4} e^{- 2 t} - \frac{17 t}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) + 17 = 0

Solve linear ODE:

y = 1 + \frac{15}{4} - \frac{15}{4} e^{- 2 t} - \frac{17 t}{2}

y = 1 + \frac{15}{4} - \frac{15}{4} e^{- 2 t} - \frac{17 t}{2}

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

y^{^{''}} + 40 y^{^{'}} + 400 y = 100 sin (10 t)

y^{^{''}} + 7 y = x

y^{^{''}} + 16 y = 8 sin (4 t)

y^{''} + y = 8 H (t - 5)

What is the solution for y^{''}+2y^'+17=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+2y^'+17=0,y(0)=1,y^'(0)=-1 is y=1+15/4-15/4 e^{-2t}-(17t)/2