What is the solution for y'=y(3-y)-9/4 ?

The solution for y'=y(3-y)-9/4 is y= 1/(t+c_{1)}+(3t)/(2(t+c_{1))}+(3c_{1})/(2(t+c_{1))}

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y'=y(3-y)-9/4

y^{^{'}} = y (3 - y) - \frac{9}{4}

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

Solution steps

y^{'} (t) = y (3 - y) - \frac{9}{4}

Solve separable ODE:

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

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

y = \frac{( 3 x ^{2} + 4 ) ( 2 x + 1 )}{5 x - 9}

f (z) = \frac{z - 8}{z + 8}

x \to 3 lim (\frac{5}{x ^{2} - 9})

x \to 0 + lim (x^{4 x})

What is the solution for y'=y(3-y)-9/4 ?
The solution for y'=y(3-y)-9/4 is y= 1/(t+c_{1)}+(3t)/(2(t+c_{1))}+(3c_{1})/(2(t+c_{1))}