What is the solution for (dy)/(dx)=(x-y)/(x+2y),y(0)=1 ?

The solution for (dy)/(dx)=(x-y)/(x+2y),y(0)=1 is y=(-x+sqrt(3x^2+4))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(x-y)/(x+2y),y(0)=1

\frac{d y}{d x} = \frac{x - y}{x + 2 y}, y (0) = 1

y = \frac{- x + 3 x ^{2} + 4}{2}

Solution steps

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

Solve exact ODE:

y = \frac{- x + 3 x ^{2} + 4}{2}

y = \frac{- x + 3 x ^{2} + 4}{2}

\int_{0}^{2} \frac{1}{( x + 1 ) ^{4}} d x

f (x) = (arccot (x))^{5}

\int (x + 1) (x - 2) d x

f (x) = \frac{1 - 5 x}{4 + x}

\frac{\partial}{\partial x} (π^{7})

What is the solution for (dy)/(dx)=(x-y)/(x+2y),y(0)=1 ?
The solution for (dy)/(dx)=(x-y)/(x+2y),y(0)=1 is y=(-x+sqrt(3x^2+4))/2