xydx+\{\frac{1}{2}x^{2}(1-y \ln y-y)\} dy=0

David Lewis 2021-12-31 Answered
xydx+{12x2(1ylnyy)}dy=0
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Expert Answer

Gerald Lopez
Answered 2022-01-01 Author has 29 answers
Step 1
The differential equation is xydx+{12x2(1ylnyy)}dy=0
Rewrite the equation in the form of separable equation as follows.
xydx+{12x2(1ylnyy)}dy=0
xy+{12x2(1ylnyy)}dydx=0 (Divide by dx on both sides)
12x2(1ylnyy)dydx=xy
(1ylnyy)ydydx=2xx2
(1ylnyy)ydydx=2x
(1ylnyy)ydy=2xdx
Step 2
Now, integrate both sides of the obtained equation and find the solution of the differential equation.
(1ylnyy)ydy=2xdx
1ydyln(y)dydy=21xdx
ln(y)yln(y)+yy=2ln(x)+C
ln(y)yln(y)+2ln(x)=C
yln(y)+ln(y)+ln(x2)=C
Donald Cheek
Answered 2022-01-02 Author has 41 answers
Step 1
The given differntial equation is xydx+(12x2(1ylnyy))dy=0.
It can be seen that it is the first order seperable differntial equation.
Now seperate the variables.
(ln(y)1+1y)dy=2xdx
Step 2
Integrate both sides.
(ln(y)1+1y)dy=2xdx
(ln(y))dy+(1+1y)dy=2xdx
(yln(y)1dy)+(1)dy+1ydy=2xdx using by part {u=lny,v=1}
ylny+1ydy=2xdx
ylny+lny=2lnx+c1
ylny+lny2lnx=c1
ln(y)y+ln(x2y)=c1 [lnx+lny=ln(xy)]
Answer: The solution of the diffrential equation is ln(y)y+ln(x2y)=c1
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user_27qwe
Answered 2022-01-05 Author has 229 answers

Given: xydx+{12x2(1ylnyy)}dy=0
The eqn can be written as
xydx=12x2(1ylnyy)dy
or 2x3dx=(1ylny1)dy
2x3dx=(1+lny1y)dy
Integrating both sides
2x3dx=(1+lny1y)dy+c
x42=y+lnydylny+c (1)
Here lny=1lny=ylny1yydy
=ylnyy(1)
x22=y+ylnyylny+c
or x22=(y1)lny+c

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