What is the solution for y^'=((x-ysin(2x)))/(sin^2(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=((x-ysin(2x)))/(sin^2(x))

y^{^{'}} = \frac{( x - y sin ( 2 x ) )}{sin ^{2} ( x )}

y = \frac{x ^{2}}{2 sin ^{2} ( x )} + \frac{c _{1}}{sin ^{2} ( x )}

Solution steps

y^{'} (x) = \frac{( x - y sin ( 2 x ) )}{sin ^{2} ( x )}

Solve linear ODE:

y = \frac{x ^{2}}{2 sin ^{2} ( x )} + \frac{c _{1}}{sin ^{2} ( x )}

y = \frac{x ^{2}}{2 sin ^{2} ( x )} + \frac{c _{1}}{sin ^{2} ( x )}

n = 1 \sum \infty \frac{1}{n \cdot 3 ^{n}}

3 e^{- 7 x}

x \to - \infty lim (3 - x)

\frac{4}{2 s ^{2} + 2 s + 1}

y = e^{4} x^{2} + 1

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