What is the solution for yy^'=(y^2+1)sin(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

yy^'=(y^2+1)sin(x)

y y^{^{'}} = (y^{2} + 1) sin (x)

y = e^{- 2 c o s (x) + c_{1}} - 1, y = - e^{- 2 c o s (x) + c_{1}} - 1

Solution steps

y y^{'} (x) = (y^{2} + 1) sin (x)

Solve separable ODE:

y = e^{- 2 c o s (x) + c_{1}} - 1, y = - e^{- 2 c o s (x) + c_{1}} - 1

y = e^{- 2 c o s (x) + c_{1}} - 1, y = - e^{- 2 c o s (x) + c_{1}} - 1

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