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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=sin^2(x-y+1)

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

y = - arctan (x + c_{1}) - πn + x + 1

Solution steps

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

Solve using substitution:

y = - arctan (x + c_{1}) - πn + x + 1

y = - arctan (x + c_{1}) - πn + x + 1

\frac{d y}{d x} + 2 y = x

\int \frac{1}{144 - x ^{2}} d x

\int arccot (8 y) d y

\int \frac{1 - x}{1 + x} d x

\int 58

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