Let |x| = |y|exy - , , N be the solution of the differential equation x dy - y dx + xy(x dy + y dx) = 0, y(1) = 2. Then + is equal to
Answer
4
Solution
Divide by xy: x dy - y dxxy + (x dy + y dx) = 0
dyy - dxx + d(xy) = 0
|y| - |x| + xy = c
y(1) = 2 c = 2 + 2
|y|2|x| = 2 - xy 2|x| = |y|exy - 2
= 2, = 2
∴ + = 4
Answer: 4