What is the solution for y^'=y(7-y)-49/4 ?

The solution for y^'=y(7-y)-49/4 is y= 1/(t+c_{1)}+(7t)/(2(t+c_{1))}+(7c_{1})/(2(t+c_{1))}

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y^'=y(7-y)-49/4

y^{^{'}} = y (7 - y) - \frac{49}{4}

y = \frac{1}{t + c _{1}} + \frac{7 t}{2 ( t + c _{1} )} + \frac{7 c _{1}}{2 ( t + c _{1} )}

Solution steps

y^{'} (t) = y (7 - y) - \frac{49}{4}

Solve separable ODE:

y = \frac{1}{t + c _{1}} + \frac{7 t}{2 ( t + c _{1} )} + \frac{7 c _{1}}{2 ( t + c _{1} )}

y = \frac{1}{t + c _{1}} + \frac{7 t}{2 ( t + c _{1} )} + \frac{7 c _{1}}{2 ( t + c _{1} )}

\frac{d}{d x} (3 x^{3} - 5 x)

\frac{2}{x ^{2}} + \frac{3}{x ^{3}}

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

\int (8 x - 2 x) d x

\frac{d}{d x} (arctan (\frac{1}{x ^{2}}))

What is the solution for y^'=y(7-y)-49/4 ?
The solution for y^'=y(7-y)-49/4 is y= 1/(t+c_{1)}+(7t)/(2(t+c_{1))}+(7c_{1})/(2(t+c_{1))}