What is the solution for y^'=y-4y^2 ?

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

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y^'=y-4y^2

y^{^{'}} = y - 4 y^{2}

y = - \frac{e ^{t + c_{1}}}{4 ( 1 - e ^{t + c_{1}} )}

Solution steps

y^{'} (t) = y - 4 y^{2}

Solve separable ODE:

y = - \frac{e ^{t + c_{1}}}{4 ( 1 - e ^{t + c_{1}} )}

y = - \frac{e ^{t + c_{1}}}{4 ( 1 - e ^{t + c_{1}} )}

y^{^{'}} = \frac{x ^{2}}{y ( 1 + x ^{3} )}

\frac{d}{d x} (2^{c o t (5 x)})

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

\int \frac{( x - 2 ) 1 - x ^{2} + x}{1 - x ^{2}} d x

\int x sec^{2} (5 x) d x

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