What is the solution for (dy)/(dx)=(x+y+8)^2 ?

The solution for (dy)/(dx)=(x+y+8)^2 is y=tan(x+c_{1})-x-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(x+y+8)^2

\frac{d y}{d x} = (x + y + 8)^{2}

y = tan (x + c_{1}) - x - 8

Solution steps

\frac{d y}{d x} = (x + y + 8)^{2}

Solve using substitution:

y = tan (x + c_{1}) - x - 8

y = tan (x + c_{1}) - x - 8

y^{^{''}} + 2 y + y = 1

\frac{\partial}{\partial x} (\frac{x}{( x ^{2} + y ^{2} ) ^{2}})

\int x^{2} 2 - x^{3} d x

cos (x), at \frac{π}{2}

\frac{\partial}{\partial y} ((x - y) (1 - x y))

What is the solution for (dy)/(dx)=(x+y+8)^2 ?
The solution for (dy)/(dx)=(x+y+8)^2 is y=tan(x+c_{1})-x-8