What is the solution for y^'+y^2cos(x)=0,y(0)=x(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+y^2cos(x)=0,y(0)=x(x)

y^{^{'}} + y^{2} cos (x) = 0, y (0) = x (x)

y = - \frac{1}{- sin ( x ) + c _{1}}

Solution steps

y^{'} (x) + y^{2} cos (x) = 0

Solve separable ODE:

y = - \frac{1}{- sin ( x ) + c _{1}}

y = - \frac{1}{- sin ( x ) + c _{1}}

\frac{d y}{d t} = 2 y - 5

\frac{d v}{d t} = - 4 v

x^{^{'}} = x^{2} + \frac{1}{49}, x (0) = - 7

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

\frac{d x}{d t} = \frac{3}{4} - \frac{3 x}{100}

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