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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

yy^'+x=0

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

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

Solution steps

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

Solve separable ODE:

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

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

x^{\frac{3}{4}}

\int y^{2} (1 - 2 x) d y

\frac{\partial}{\partial x} (y (e^{x} - 1))

x \to 0 lim (x^{2} + 16 x + 25)

k = 0 \sum \infty \frac{2}{k !}

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