What is the solution for 2y^'=y^2-6y+9 ?

The solution for 2y^'=y^2-6y+9 is y=-2/(t+c_{1)}+(3t)/(t+c_{1)}+(3c_{1})/(t+c_{1)}

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2y^'=y^2-6y+9

2 y^{^{'}} = y^{2} - 6 y + 9

y = - \frac{2}{t + c _{1}} + \frac{3 t}{t + c _{1}} + \frac{3 c _{1}}{t + c _{1}}

Solution steps

2 y^{'} (t) = y^{2} - 6 y + 9

Solve separable ODE:

y = - \frac{2}{t + c _{1}} + \frac{3 t}{t + c _{1}} + \frac{3 c _{1}}{t + c _{1}}

y = - \frac{2}{t + c _{1}} + \frac{3 t}{t + c _{1}} + \frac{3 c _{1}}{t + c _{1}}

\frac{d v}{d t} = 6 - 8 v

\frac{d y}{d x} + y = y^{- \frac{1}{2}}

f (x) (x + 1) = f^{^{'}} (x)

\frac{d y}{d x} = \frac{( x y + 5 x - y - 5 )}{x y - 2 x + 6 y - 12}

(2 + x) y^{^{'}} = y

What is the solution for 2y^'=y^2-6y+9 ?
The solution for 2y^'=y^2-6y+9 is y=-2/(t+c_{1)}+(3t)/(t+c_{1)}+(3c_{1})/(t+c_{1)}