求微分方程式 (2x^2+3y^2 -7) x dx -(3x^2+2y^2 -8 ) y dy=0 請問一下該如呵計算?

0 = (2x^2+3y^2-7) x dx - (3x^2+2y^2-8) y dy

= (2x^2+3y^2-7)d(x^2) - (3x^2+2y^2-8)d(y^2)


2x^2 + 3y^2 = 7

3x^2 + 2y^2 = 8



=> x^2 = 2, y^2 = 1

=> x2 = u + 2, y^2 = v + 1

=> v = ut, t = v/u

=> dv = udt + tdu



0 = (2u + 3v)du - (3u+2v)dv

= (2u + 3ut)du - (3u + 2ut)(udt + tdu)

= (2 + 3t)du - (3 + 2t)(udt + tdu)

= 2(1-t^2)du - (3-2t)du

= 2du/u + (2t - 3)dt/(1-t^2)


ln(c) = ∫2du/u + ∫dt/2(t-1) - ∫5dt/2(t+1)

= ln{u^2/(t+1)^2*√[(t-1)/(t+1)]}

c = u^2/(v/u + 1)^2*√[(v/u - 1)/(v/u + 1)]

= u^4/(u+v)^2*√[(v-u)/(v+u)]

= (x^2-2)^4/(x^2+y^2-3)^2 * √[(y^2-x^2+1)/(x^2+y^2-3)]

= Answer

http://youtu.be/SBYvc5OxK8w