What is the solution for y^'= 3/(3y) ?

The solution for y^'= 3/(3y) is y=sqrt(2t+c_{1)},y=-sqrt(2t+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'= 3/(3y)

y^{^{'}} = \frac{3}{3 y}

y = 2 t + c_{1}, y = - 2 t + c_{1}

Solution steps

y^{'} (t) = \frac{3}{3 y}

Solve separable ODE:

y = 2 t + c_{1}, y = - 2 t + c_{1}

y = 2 t + c_{1}, y = - 2 t + c_{1}

\int x^{3} ln (x^{2} + 1) d x

\int \frac{8}{t ^{9}} sin (\frac{1}{t ^{8}} - 9) d t

y^{^{'}} + tan (x) y = cos^{2} (x)

\frac{\partial}{\partial x} (tan (2 x - y))

\frac{\partial}{\partial x} ((x^{2} - 2) e^{x^{2} y})

What is the solution for y^'= 3/(3y) ?
The solution for y^'= 3/(3y) is y=sqrt(2t+c_{1)},y=-sqrt(2t+c_{1)}