What is the solution for y^'=y(4-ty) ?

The solution for y^'=y(4-ty) is y=(16e^{4t})/(4e^{4t)t-e^{4t}+c_{1}}

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y^'=y(4-ty)

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

y = \frac{16 e ^{4 t}}{4 e ^{4 t} t - e ^{4 t} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{16 e ^{4 t}}{4 e ^{4 t} t - e ^{4 t} + c _{1}}

y = \frac{16 e ^{4 t}}{4 e ^{4 t} t - e ^{4 t} + c _{1}}

\frac{d}{d x} (cos^{2} (x) + sin^{2} (x))

\int x^{2} cos (8 x) d x

f (x) = \frac{56}{x ^{9}}

x \to 0 lim (x^{2} - 3)

x \to 2 lim (\frac{x + 5}{x - 2})

What is the solution for y^'=y(4-ty) ?
The solution for y^'=y(4-ty) is y=(16e^{4t})/(4e^{4t)t-e^{4t}+c_{1}}