Question

xydx + (x+1)dy = 0

Answer 1

diff eq?

Answer 2

yes

Answer 3

\[xydx+(x+1)dy\]

\[-(x+1)dy=xydx\]

\[\frac{1}{y}dy=\frac{x}{-(x+1)}dx\]

Answer 4

xy dx=-(x+1)dy
((x)/(x+1))dx=-(1/y)dy
integrate both sides and u will get the solution....

Answer 5

this is known as variable separable form.....

Answer 6

\[xydx+(x+1)dy=0\]

\[xy+(x+1)\frac{dy}{dx}=0\]

Answer 7

\[(x+1)\frac{dy}{dx}+xy=0\]

\[\frac{dy}{dx}+\frac{x}{x+1}y=0\]

Answer 8

find \[\mu\]

Answer 9

xy(x) + (x +1) (dy(x)) / (dx)
dy(x) / dx = (- xy(x)) / (x + 1)

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

(integral) ( dy(x) / dx ) / y(x) dx = (integral) - x / (x + 1)
evaluate

ln|y(x)| = c1 - x + ln|x + 1| where c1 is an arbitrary constant

solve for y(x)
y(x) = (x+1)e^(c1-x)

y(x) = ( c1(x +1) / (e^x) )