What is the solution for y^'sin^2(x)=cos(x)(1-3y) ?

The solution for y^'sin^2(x)=cos(x)(1-3y) is y=-(e^{3csc(x)-c_{1}})/3+1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'sin^2(x)=cos(x)(1-3y)

y^{^{'}} sin^{2} (x) = cos (x) (1 - 3 y)

y = - \frac{e ^{3 c s c (x) - c_{1}}}{3} + \frac{1}{3}

Solution steps

y^{'} (x) sin^{2} (x) = cos (x) (1 - 3 y)

Solve separable ODE:

y = - \frac{e ^{3 c s c (x) - c_{1}}}{3} + \frac{1}{3}

y = - \frac{e ^{3 c s c (x) - c_{1}}}{3} + \frac{1}{3}

What is the solution for y^'sin^2(x)=cos(x)(1-3y) ?
The solution for y^'sin^2(x)=cos(x)(1-3y) is y=-(e^{3csc(x)-c_{1}})/3+1/3