這種solve first order differential equation方法的原理是甚麼？

This is called Exact Differential Equations

I do it once again

solve the inital value problem
(y cos(xy) + x) dx + (y + x cos(xy)) dy = 0, y(0) = 1

∫ y cos(xy) + x dx
=sin(xy)+x^2/2+f(y)

d/dy[sin(xy)+x^2/2+f(y)]
=xcos(xy)+f'(y)

So f'(y)=y

f(y)=y^2/2+k

The F(x,y)=sin(xy)+x^2/2+y^2/2+k

Now (y cos(xy) + x) dx + (y + x cos(xy)) dy = 0 becomes

sin(xy)+x^2/2+y^2/2+k=c

or sin(xy)+x^2/2+y^2/2=c1 where c1 is a constant

y(0)=1

c1=1/2

So the answer is sin(xy)+x^2/2+y^2/2=1/2

2008-03-24 00:06:47 補充：
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Exact Differential Equation

http://www.sosmath.com/diffeq/first/exact/exact.html