Question

the curve satisfies the differential equation (x^2-y^2)dx+2xydy=0 and passes through pt (1,1) is

Answer 1

If x = r sec & y = r tan then x dx  y dy = r dr and x dy  y dx = r2 secd .
TYPE2 : dy
dx
= f (ax + by + c) , b  0.
To solve this , substitute t = ax + by + c. Then the equation reduces to separable type in the
variable t and x which can be solved.