Question

find the solution of differential equation (dy)/(dx)=(x+y+2)/2 (x+y)+3​

Answer 1

Step-by-step explanation:

$\dfrac{dy}{dx} = \dfrac{x + y + 2}{2(x + y) + 3} \\ let \\ \: \: (x + y) = v \\ differentiating \: with \: respect \: to \: x \\ 1 + \dfrac{dy}{dx} = \dfrac{dv}{dx} \\ \dfrac{dy}{dx} = \dfrac{dv}{dx} - 1 \\ now \\ \dfrac{dv}{dx} - 1 = \dfrac{v + 2}{2v + 3} \\ \dfrac{dv}{dx} = \dfrac{v + 2}{2v + 3} + 1 \\ \dfrac{dv}{dx} = \dfrac{3v + 5}{2v + 3} \\ \displaystyle \int \dfrac{2v + 3}{3v + 5} \: dv = \displaystyle \int dx \: + \: c \\ 2 \displaystyle \int \dfrac{v + \frac{3}{2} }{3v + 5} \: dv = \: x \: + \: c \\ \dfrac{2}{3} \displaystyle \int \: \dfrac{3v + \frac{9}{2} }{3v + 5} dv= \: x \: + \: c \\ \dfrac{2}{3} \displaystyle \int \dfrac{(3v + 5) - 5 + \frac{9}{2} }{3v + 5} \: dv = \: x \: + \: c \\ \dfrac{2}{3} \displaystyle \int( \: \dfrac{3v + 5}{3v + 5} - \dfrac{ \frac{1}{2} }{3v + 5} ) \: dv \: = \: x \: + \: c \\ \dfrac{2}{3} \displaystyle \int( \: 1 - \dfrac{1}{2 \: (3v + 5)} ) \: dv \: = x \: + \: c \\ \dfrac{2}{3} ( \: v) - \frac{2}{3} \: \: \dfrac{1}{2} \: \dfrac{1}{3} \: log \: (3v + 5) = x + c \\ \dfrac{2}{3} (x + y) - \dfrac{1}{9} \: log (3 \: (x + y) + 5) = x + c \\ \dfrac{2}{3} (x + y) - \dfrac{1}{9} log(3x + 3y + 5) = x + c$