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

The solution for sin(x)y^'=y+2x is y=(integral of 2xcsc(x)cot(x/2))/(cot(x/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{\int 2 x csc ( x ) cot ( \frac{x}{2} ) d x}{cot ( \frac{x}{2} )}

Solution steps

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

Solve linear ODE:

y = \frac{\int 2 x csc ( x ) cot ( \frac{x}{2} ) d x}{cot ( \frac{x}{2} )}

y = \frac{\int 2 x csc ( x ) cot ( \frac{x}{2} ) d x}{cot ( \frac{x}{2} )}

\int (cos (x) + \frac{1}{9} x) d x

x \to 1 lim (\frac{2 - 7 x}{x + 6})

5 lo g_{10} (x)

y^{^{''}} - 7 y^{^{'}} = 49 t, y (0) = 7, y^{^{'}} (0) = 0

x \to 0 lim (\frac{1}{x ^{2}})

What is the solution for sin(x)y^'=y+2x ?
The solution for sin(x)y^'=y+2x is y=(integral of 2xcsc(x)cot(x/2))/(cot(x/2))