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Solve: (3xy+y) dx + (x + xy) dy = 0
Please give the steps after integration begins.
there is no need for long integration method just open the brackets
3xydx + ydx + xdy+xydy = 0
ydx+xdy=-(3xydx+xydy)
ydx+xdy/xy = 3dx+dy
now see the lhs this is the differential of logxy (u have to memorise some certain forms like these plus u can check it by differentiating logxy)
now integrating it becomes
logxy = 3x + y + c
can covert to a required form if needed
solution:
=>(3xy+y)dx+(x+xy) dy=0
=>(3x+1) y dx +(1+y) x dy=0
=>(3x+1) y dx =- (1+y) x dy
=>((3x+1)/ x) dx=-((1+y)/y) dy
=>(3+1/x)dx=(-1/y -1)dy
then intigrate by parts ....
=>3x+lnx=-lny-y
answer is 3x+y+lnxy=0.
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