What is the solution for 3x^2y^2dx+2x^3ydy=0,y(1)=2 ?

The solution for 3x^2y^2dx+2x^3ydy=0,y(1)=2 is y=(c_{1})/(x^{3/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3x^2y^2dx+2x^3ydy=0,y(1)=2

3 x^{2} y^{2} d x + 2 x^{3} y d y = 0, y (1) = 2

y = \frac{c _{1}}{x ^{\frac{3}{2}}}

Solution steps

3 x^{2} y^{2} d x + 2 x^{3} y d y = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{\frac{3}{2}}}

y = \frac{c _{1}}{x ^{\frac{3}{2}}}

(\frac{d v}{d t} + k v (t)) e^{k t} = g e^{k t}

y^{^{'}} (t) = (1 + y) (3 - y), y (0) = 4

\frac{d y}{d x} = 0.04 - \frac{y}{\frac{4320}{7}}

\frac{d y}{d x} = (1 + 2 x) y

f (x) = f^{^{'}} (x)

What is the solution for 3x^2y^2dx+2x^3ydy=0,y(1)=2 ?
The solution for 3x^2y^2dx+2x^3ydy=0,y(1)=2 is y=(c_{1})/(x^{3/2)}