# (2y+xy)dx+2xdy=0

Question
Differential equations
$$\displaystyle{\left({2}{y}+{x}{y}\right)}{\left.{d}{x}\right.}+{2}{x}{\left.{d}{y}\right.}={0}$$

2020-12-17
Here the Differential equations is given by
$$\displaystyle{\left({2}{y}+{x}{y}\right)}{\left.{d}{x}\right.}+{2}{x}{\left.{d}{y}\right.}={0}.$$
This can be written as
PSKy(2+x)dx+2xdy=0  \Rightarrow  \frac{2+x}{x}dx+\frac{2}{y}dy=0   \Rightarrow  (\frac{2}{x}+1)dx+\frac{2}{y}dy=0   \Rightarrow  d(2\ln x+x)+d(2\ln y)=0   \Rightarrow  d(2\ln x+x+2\ln y)=0  \Rightarrow  ∫d(2\ln x+x+2\ln y)=c   \Rightarrow  2\ln x+x+2\ln y=c   \Rightarrow \ln y=(c-2\ln x-x)/2   \Rightarrow  y=e^{\frac{c-2\ln x-x}{2}}ZSK
where cc is a arbitrary constants.

### Relevant Questions

Solve the differential equations
(1) $$\displaystyle{x}{y}'-{2}{y}={x}^{{3}}{e}^{{x}}$$
(2) $$\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{2}}{x}={0}$$
If $$\displaystyle{x}{y}+{6}{e}^{{{y}}}={6}{e}$$, find the value of y" at the point where x = 0.
Solve this equation pls $$\displaystyle{y}'+{x}{y}={e}^{{x}}$$
$$\displaystyle{y}{\left({0}\right)}={1}$$
Solve the initial value problem.
$$\displaystyle{4}{x}^{{2}}{y}{''}+{17}{y}={0},{y}{\left({1}\right)}=-{1},{y}'{\left({1}\right)}=-\frac{{1}}{{2}}$$
Solve the differential equation $$2y"+20y'+51y=0, y(0)=2$$
$$y'(0)=0$$
Q. 1# $$(a−x)dy+(a+y)dx=0(a−x)dy+(a+y)dx=0$$
Q. 1# $$\displaystyle{\left({a}−{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}{\left({a}−{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$
Solve. $$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+\frac{{3}}{{x}}{y}={27}{y}^{{\frac{{1}}{{3}}}}{1}{n}{\left({x}\right)},{x}{>}{0}$$
Find the differential dy for the given values of x and dx. $$y=\frac{e^x}{10},x=0,dx=0.1$$