What is the solution for y^'=(x^3)/(3e^{y+1)} ?

The solution for y^'=(x^3)/(3e^{y+1)} is y=ln((x^4)/(12)+c_{1})-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(x^3)/(3e^{y+1)}

y^{^{'}} = \frac{x ^{3}}{3 e ^{y + 1}}

y = ln (\frac{x ^{4}}{12} + c_{1}) - 1

Solution steps

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

Solve separable ODE:

y = ln (\frac{x ^{4}}{12} + c_{1}) - 1

y = ln (\frac{x ^{4}}{12} + c_{1}) - 1

(ln (y))^{3} (\frac{d y}{d x}) = x^{3} y, y (1) = e^{2}

\frac{d y}{d x} = k 2 y + a

(y^{^{'}}) = e^{x} e^{y}

\frac{1}{x} d x + \frac{1}{2 y} d y = 0

(t^{2} + 1) y^{^{'}} = sin (t^{2} + 3 cos (t)) y

What is the solution for y^'=(x^3)/(3e^{y+1)} ?
The solution for y^'=(x^3)/(3e^{y+1)} is y=ln((x^4)/(12)+c_{1})-1