
Yefim S. answered 03/04/21
Math Tutor with Experience
dy/dx = y/x - a(1 - y/x)1/2; y(a) = 0
(a) This is Homoginius equation, because right side f(x, y ) = y/x - a(1 - y/x)1/2 = f(kx, ky)
(b) L<et y = vx, where v is new function of x. Then dy/dx = v + xdv/dx and we sabstitute this in our O)DE:
v + xdv/dx = v - a(1 - v)1/2; dv/(1 - v)1/2 = - adx/x. We integrate both sides: ∫dv/(1 - v)1/2 = ∫-adx/x
-2(1 - v)1/2 = -alnIxI - C; Let put back v = y/x; 2(1 - y/x)1/2 = alnIxI + C. If x = a y(a) = 0; 2 = alnIaI + C;
C = 2 - alnIaI
2(1 - y/x)1/2 = alnIxI + 2 - alnIaI; 4(1 - y/x) = (alnIx/aI + 2)2; y = x(1 - (alnIx/aI + 2)2/