Multiply the original DE by xy:
xy2(1+x2y4+1−−−−−−−√)dx+2x2ydy=0(1)
Let v=xy2, so that dv=y2dx+2xydy. Then (1) becomes
x(y2dx+2xydy)+xy2x2y4+1−−−−−−−√dxxdv+vv2+1−−−−−√dx=0=0
This final equation is easily recognized as separable:
dxxln|x|+CKxvKx2y2−1K2x4y4−2Kx2y2y2=−dvvv2+1−−−−−√=ln∣∣∣v2+1−−−−−√+1v∣∣∣=v2+1−−−−−√+1=x2y4+1−−−−−−−√=x2y4=2KK2x2−1integrate both sides
I'm calculated and I figure that it's graphing
Solution,
y = -x-4......(i)
3x + y =- 16.......(ii)
Now,
Substituting the value of y in equation (ii)
or, 3x + (-x-4)= -16
or,3x - x - 4 = -16
or, 2x= -16 +4
or, x = -12/2
: x = -6
Then,
Substituting the value of x in (i) equation;
or,y= -(-6) - 4
or,y= 6 - 4
: y =2
