What is the solution for yy^'+6x=0 ?

The solution for yy^'+6x=0 is y=sqrt(-6x^2+c_{1)},y=-sqrt(-6x^2+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

yy^'+6x=0

y y^{^{'}} + 6 x = 0

y = - 6 x^{2} + c_{1}, y = - - 6 x^{2} + c_{1}

Solution steps

y y^{'} (x) + 6 x = 0

Solve separable ODE:

y = - 6 x^{2} + c_{1}, y = - - 6 x^{2} + c_{1}

y = - 6 x^{2} + c_{1}, y = - - 6 x^{2} + c_{1}

\int x^{2} (x^{3} - 4)^{4} d x

\int 3^{- x} cos (x) d x

\int 1 + \frac{1}{x} d x

\frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

\frac{d y}{d x} + \frac{5 x ^{2}}{e ^{3 x^{3}}} y = 0

What is the solution for yy^'+6x=0 ?
The solution for yy^'+6x=0 is y=sqrt(-6x^2+c_{1)},y=-sqrt(-6x^2+c_{1)}