⑩天掌 Teaching Plan on Advanced Mathematics 例2求方程f(x)y+g(x)x=0通解 解令u=x,则d=xd+ytx, ∫(u)yx+g(u)x du-ydx 0 lf(u)-gul-d+gudu=0, dx gu x ulf(u-al du=0, 通解为In|x g u u∫(u)-g() tianjin Polytechnic univerityTianjin Polytechnic University Teaching Plan on Advanced Mathematics u = xy, du = xdy + ydx, ( ) ( ) = 0, − +  x du ydx f u ydx g u x [ ( )− ( )] dx + g(u)du = 0, x u f u g u 0, [ ( ) ( )] ( ) = − + du u f u g u g u x dx 通解为 解 求方程 f (xy)ydx + g(xy)xdy = 0 通解. . [ ( ) ( )] ( ) ln | | du C u f u g u g u x = − + 令 则 例2