Question

y(1+x)dx+x(1+y)dy=0​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The general solution is $ye^y=\dfrac{c}{xe^x}$.

Step-by-step explanation:

The given equation is

$y(1+x)dx+x(1+y)dy=0$

Separate the variables.

$x(1+y)dy=-y(1+x)dx$

$\dfrac{(1+y)}{y}dy=-\dfrac{(1+x)}{x}dx$

$(\dfrac{1}{y}+1)dy=-(\dfrac{1}{x}+1)dx$

Integrate both sides.

$\int (\dfrac{1}{y}+1)dy=-\int (\dfrac{1}{x}+1)dx$

$\ln y+y=-(\ln x+x)+\lnc$

$\ln y+\n e^y=-(\ln x+\ln e^x)+\ln c$$\ln ye^y=\ln(\dfrac{c}{xe^x})$

$ye^y=\dfrac{c}{xe^x}$

Therefore the general solution is $ye^y=\dfrac{c}{xe^x}$.

#Learn more

Y sin2x dx -(1+y² + cos²x) dy=0.​

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