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

The solution for y^'=(x+y-2)^2,y(0)=0 is y=tan(x-1.10714…+pin)-x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = tan (x - 1.10714 \dots + πn) - x + 2

Solution steps

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

Solve using substitution:

y = tan (x - 1.10714 \dots + πn) - x + 2

y = tan (x - 1.10714 \dots + πn) - x + 2

2 cos (π t)

\frac{\partial}{\partial y} ((x - 5)^{\frac{1}{4}} y^{\frac{1}{4}})

\frac{d}{d x} (x^{x + 1})

\frac{d}{d x} (- e^{- x y})

\int_{1}^{\infty} (e^{- x}) d x

What is the solution for y^'=(x+y-2)^2,y(0)=0 ?
The solution for y^'=(x+y-2)^2,y(0)=0 is y=tan(x-1.10714…+pin)-x+2