What is the solution for y^'= t/(1+t^2)*e^y ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'= t/(1+t^2)*e^y

y^{^{'}} = \frac{t}{1 + t ^{2}} \cdot e^{y}

y = - ln (- \frac{1}{2} ln (1 + t^{2}) - c_{1})

Solution steps

y^{'} (t) = \frac{t}{1 + t ^{2}} \cdot e^{y}

Solve separable ODE:

y = - ln (- \frac{1}{2} ln (1 + t^{2}) - c_{1})

y = - ln (- \frac{1}{2} ln (1 + t^{2}) - c_{1})

n \to \infty lim (\frac{x}{n})

\frac{d}{d r} (\frac{1}{r ^{2}})

\frac{1}{3} x^{- \frac{2}{3}}

\int sin^{3} (v) cos^{8} (v) d v

\frac{d}{d n} ((n + 1)^{n})

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