II. (x + y) dy + (x-y) dx = 0; y = | x = |
Hey student,
Since this is a homogeneous differential equation, let y = xvy = xv . As dy = vdx + xdvdy = vdx + xdv , we obtain
(x−vx)dx + (x+vx)(vdx+xdv) = 0
(x−vx)dx + (x+vx)(vdx+xdv) = 0.
This simplifies to
(1+v2)dx + x(1+v)dv = 0
(1+v2)dx + x(1+v)dv=0
Dividing both sides by x(1+v2)x(1+v2) :
dxx+1+v1+v2dv=0
dxx+1+v1+v2dv=0
Now, we can integrate both sides:
ln|x| + arctanv + 12ln(1+v2) = A
ln|x| + arctanv + 12ln(1+v2) = A.
To make this solution look a little nicer, multiply both sides by 2 (and let C=2AC=2A ):
2ln|x|+ 2arctanv + ln(1+v2) = C
2ln|x|+ 2arctanv + ln(1+v2) = C.
Rewrite this in terms of yy :
ln(x2) + 2arctan(yx) + ln(1+y2x2) = C
ln(x2) + 2arctan(yx) + ln(1+y2x2) = C
Finally, we use the product rule for logarithms:
2arctan(yx) + ln(x2+y2) = C
I hope you find it useful.
