What is the solution for y^{''}+6y+34=0,y(0)=3 ?

The solution for y^{''}+6y+34=0,y(0)=3 is y= 26/3 cos(sqrt(6)t)+c_{2}sin(sqrt(6)t)-17/3

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y^{''}+6y+34=0,y(0)=3

y^{^{''}} + 6 y + 34 = 0, y (0) = 3

y = \frac{26}{3} cos (6 t) + c_{2} sin (6 t) - \frac{17}{3}

Solution steps

y^{''} (t) + 6 y + 34 = 0

Solve linear ODE:

y = \frac{26}{3} cos (6 t) + c_{2} sin (6 t) - \frac{17}{3}

y = \frac{26}{3} cos (6 t) + c_{2} sin (6 t) - \frac{17}{3}

y^{^{''}} + 2 y^{^{'}} - 3 y = 3 e^{2 t}

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

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + e - x

y^{^{''}} + 2 y = t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 8

y^{^{''}} - y = - 15 t + 4

What is the solution for y^{''}+6y+34=0,y(0)=3 ?
The solution for y^{''}+6y+34=0,y(0)=3 is y= 26/3 cos(sqrt(6)t)+c_{2}sin(sqrt(6)t)-17/3