What is the solution for y^'=y(5-y)-25/4 ,y(0)=a ?

The solution for y^'=y(5-y)-25/4 ,y(0)=a is y=(5at)/(2at-5t+2)-(25t)/(2(2at-5t+2))+(2a)/(2at-5t+2)

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y^'=y(5-y)-25/4 ,y(0)=a

y^{^{'}} = y (5 - y) - \frac{25}{4}, y (0) = a

y = \frac{5 a t}{2 a t - 5 t + 2} - \frac{25 t}{2 ( 2 a t - 5 t + 2 )} + \frac{2 a}{2 a t - 5 t + 2}

Solution steps

y^{'} (t) = y (5 - y) - \frac{25}{4}

Solve separable ODE:

y = \frac{5 a t}{2 a t - 5 t + 2} - \frac{25 t}{2 ( 2 a t - 5 t + 2 )} + \frac{2 a}{2 a t - 5 t + 2}

y = \frac{5 a t}{2 a t - 5 t + 2} - \frac{25 t}{2 ( 2 a t - 5 t + 2 )} + \frac{2 a}{2 a t - 5 t + 2}

a^{2} (d y - d x) = x^{2} d y + y^{2} d x

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

sin (3 x) d x + 2 y cos (3 x) d y = 0

y^{^{'}} tan (x) = y

\frac{d y}{d x} = \frac{e ^{x}}{2 y}

What is the solution for y^'=y(5-y)-25/4 ,y(0)=a ?
The solution for y^'=y(5-y)-25/4 ,y(0)=a is y=(5at)/(2at-5t+2)-(25t)/(2(2at-5t+2))+(2a)/(2at-5t+2)