What is the solution for sqrt(1-x^2)y^'-sqrt(1-y^2)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(1-x^2)y^'-sqrt(1-y^2)=0

1 - x^{2} y^{^{'}} - 1 - y^{2} = 0

y = sin (arcsin (x) + c_{1})

Solution steps

1 - x^{2} y^{'} (x) - 1 - y^{2} = 0

Solve separable ODE:

y = sin (arcsin (x) + c_{1})

y = sin (arcsin (x) + c_{1})

\frac{\partial}{\partial x} (7)

f (x) = ln (x^{2} + 9)

\frac{1}{x} + tan (x)

\int \frac{1}{- 8 x - 8 x} d x

f (x) = x^{3} - 4 x, at x = - 2

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