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} )}

\frac{d y}{d x} = 5 y e^{x + 9}

\int cot^{3} (x) csc (x) d x

\int x ln (4 x) d x

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

y^{^{'}} + \frac{y}{x} - y = 0

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))}