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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(x)y^'+ycos(x)=y^2cos(x)

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

y = \frac{1 + e ^{c_{1}} sin ^{\frac{1}{2}} ( x )}{1 - e ^{2 c_{1}} sin ( x )}

Solution steps

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

Solve separable ODE:

y = \frac{1 + e ^{c_{1}} sin ^{\frac{1}{2}} ( x )}{1 - e ^{2 c_{1}} sin ( x )}

y = \frac{1 + e ^{c_{1}} sin ^{\frac{1}{2}} ( x )}{1 - e ^{2 c_{1}} sin ( x )}

\frac{\partial}{\partial s} (r sin (t) sin (s))

\frac{\partial}{\partial x} (\frac{4 x - y}{4 x + y})

\frac{d y}{d x} = y^{- 1} x^{- 3}

y^{^{'}} = \frac{6 y ^{2}}{x ^{2}}, y (- 2) = 5

\frac{d}{d x} (\frac{4}{3} tan (x))

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