What is the solution for y^'=((xy)^2)/8+y^2 ?

The solution for y^'=((xy)^2)/8+y^2 is y=-(24)/(x^3+24x+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=((xy)^2)/8+y^2

y^{^{'}} = \frac{( x y ) ^{2}}{8} + y^{2}

y = - \frac{24}{x ^{3} + 24 x + c _{1}}

Solution steps

y^{'} (x) = \frac{( x y ) ^{2}}{8} + y^{2}

Solve separable ODE:

y = - \frac{24}{x ^{3} + 24 x + c _{1}}

y = - \frac{24}{x ^{3} + 24 x + c _{1}}

\frac{d x}{d t} = 75 - \frac{x}{100}

\frac{d y}{d x} - \frac{2 x}{y} = 0

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

\frac{d V}{d t} = - 0.4 V

(1 - cos (x)) \frac{d y}{d x} = y sin (x)

What is the solution for y^'=((xy)^2)/8+y^2 ?
The solution for y^'=((xy)^2)/8+y^2 is y=-(24)/(x^3+24x+c_{1)}