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2 votes
X×cos
(y)/(x)× (ydx +xdy)= y×sin
(y)/(x)×(xdy- ydx)

User Plesiv
by
4.7k points

1 Answer

2 votes

Looks like you're given

x cos(y/x) (y dx + x dy) = y sin(y/x) (x dy - y dx)

Like with your other equation, multiply by 1/x² on both sides:

cos(y/x) (y/x dx + dy) = y/x sin(y/x) (dy - y/x dx)

Then substitute z = y/x again, so y = xz and dy = x dz + z dx. You end up with a separable equation.

cos(z) (z dx + x dz + z dx) = z sin(z) (x dz + z dx - z dx)

cos(z) (2z dx + x dz) = xz sin(z) dz

2z cos(z) dx = x (z sin(z) - cos(z)) dz

(z sin(z) - cos(z)) / (z cos(z)) dz = 2/x dx

(tan(z) - 1/z) dz = 2/x dx

Integrate both sides:

∫(tan(z) - 1/z) dz = ∫ 2/x dx

-ln|cos(z)| - ln|z| = 2 ln|x| + C

ln|cos(z)| + ln|z| = -2 ln|x| + C

ln|z cos(z)| = ln(1/x²) + C

z cos(z) = C/x²

Solve in terms of y :

(y/x) cos(y/x) = C/x²

y cos(y/x) = C/x

User Macaubas
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