What is the solution for y^'=-(4x)/y ?

The solution for y^'=-(4x)/y is y=sqrt(-4x^2+c_{1)},y=-sqrt(-4x^2+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-(4x)/y

y^{^{'}} = - \frac{4 x}{y}

y = - 4 x^{2} + c_{1}, y = - - 4 x^{2} + c_{1}

Solution steps

y^{'} (x) = - \frac{4 x}{y}

Solve separable ODE:

y = - 4 x^{2} + c_{1}, y = - - 4 x^{2} + c_{1}

y = - 4 x^{2} + c_{1}, y = - - 4 x^{2} + c_{1}

n = 0 \sum \infty \frac{8}{n !}

y^{^{'}} = 1 - y^{2}

\int x + x^{2} d x

\int 5 x^{3} d x

\frac{d}{d x} ((1 + sin (x)) cos (2 x))

What is the solution for y^'=-(4x)/y ?
The solution for y^'=-(4x)/y is y=sqrt(-4x^2+c_{1)},y=-sqrt(-4x^2+c_{1)}