What is the solution for y^'=y(2-ty) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=y(2-ty)

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

y = \frac{4 e ^{2 t}}{2 e ^{2 t} t - e ^{2 t} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{4 e ^{2 t}}{2 e ^{2 t} t - e ^{2 t} + c _{1}}

y = \frac{4 e ^{2 t}}{2 e ^{2 t} t - e ^{2 t} + c _{1}}

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