What is the solution for y^{''}+(y^')2+9=0 ?

The solution for y^{''}+(y^')2+9=0 is y=c_{1}+c_{2}e^{-2t}-(9t)/2

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y^{''}+(y^')2+9=0

y^{^{''}} + (y^{^{'}}) 2 + 9 = 0

y = c_{1} + c_{2} e^{- 2 t} - \frac{9 t}{2}

Solution steps

y^{''} (t) + (y^{'} (t)) 2 + 9 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} - \frac{9 t}{2}

y = c_{1} + c_{2} e^{- 2 t} - \frac{9 t}{2}

y^{^{''}} - y^{^{'}} = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = tan (2 t)

y^{^{''}} + 5 y^{^{'}} + 6 y = sin (2 x)

y^{^{''}} + 3 y^{^{'}} + 5 y = sin (t), y^{^{'}} (0) = 0, y (0) = 1

y^{^{''}} + (4 + 1) y^{^{'}} = - 4 y + 7 e^{- x}

What is the solution for y^{''}+(y^')2+9=0 ?
The solution for y^{''}+(y^')2+9=0 is y=c_{1}+c_{2}e^{-2t}-(9t)/2