6.4.26.4.46.3.16.3.26.4.1刘维尔公式6.4.3例3求解定解问题1 + (y')?= 2yy"y(0) = 1, y(0) = 0解这是不显含α的二阶方程令p=y,则y"=pd方程化为dp1 + p = 2yP dy即,dy2pdp1+py积分得In(1+p2)=lnlyl +Co因为y(o)=1>0,所以y不能取负值因此1+p2= Ciy返回全屏关闭退出II6/37
6.3.1 6.3.2 6.4.1 4úª 6.4.2 6.4.3 6.4.4 ~ 3 ¦)½)¯K 1 + (y 0 ) 2 = 2yy00 y(0) = 1, y0 (0) = 0 ) ù´Øw¹ x §. - p = y 0 , K y 00 = p dp dy. §z 1 + p 2 = 2yp dp dy . =, 2p 1 + p2 dp = dy y . È© ln(1 + p 2 ) = ln |y| + C0. Ï y(0) = 1 > 0, ¤± y ØUK. Ïd 1 + p 2 y = C1 6/37 kJ Ik J I £ ¶ '4 òÑ
6.3.16.3.26.4.1刘维尔公式6.4.26.4.36.4.4其中Ci是正常数.根据条件y(o)=1,y(o)=0可知Ci=1.因此1 +(y)?= y,即,dr两边积分,得±2Vy-1=α+C2再根据y(0)=1,y(0)=0可知C2=0.因此±2Vy-1=c.即,+1cy4返回全屏关闭退出?7/37
6.3.1 6.3.2 6.4.1 4úª 6.4.2 6.4.3 6.4.4 Ù¥ C1 ´~ê. â^ y(0) = 1, y0 (0) = 0 C1 = 1. Ïd 1 + (y 0 ) 2 = y, =, ± dy √ y − 1 = dx. ü>È©, ±2 p y − 1 = x + C2. 2â y(0) = 1, y0 (0) = 0 C2 = 0. Ïd ±2 p y − 1 = x. =, y = 1 4 x 2 + 1. 7/37 kJ Ik J I £ ¶ '4 òÑ
6.3.16.4.1刘维尔公式6.4.26.4.36.4.46.3.2例4求方程y"- e2y = 0满足初始条件 y(0)= 0, y'(0) = 1 的特解解这是不显含α的二阶方程.令p=y,则y"=p.代入方程后,得dpe2y = 0.pdy分离变量,并积分,得出1.1-e2y + C1.D22由 y(0) = 0, p(0) = y'(0) = 1, 得 Ci = 0. 故 p2 = e2y, 所以dyey.dac分离变量,再积分,得出-e-y =c+C.由 y(0)= 0, 得 C = -1, 故所求特解为 1-e-y = c.I返回全屏关闭退出8/37
6.3.1 6.3.2 6.4.1 4úª 6.4.2 6.4.3 6.4.4 ~ 4 ¦§ y 00 − e 2y = 0 ÷vЩ^ y(0) = 0, y0 (0) = 1 A). ) ù´Øw¹ x §. - p = y 0 , K y 00 = p dp dy. \§, p dp dy − e 2y = 0. ©lCþ, ¿È©, Ñ 1 2 p 2 = 1 2 e 2y + C1. d y(0) = 0, p(0) = y 0 (0) = 1, C1 = 0. p 2 = e 2y , ¤± dy dx = e y . ©lCþ, 2È©, Ñ −e −y = x + C. d y(0) = 0, C = −1, ¤¦A) 1 − e −y = x. 8/37 kJ Ik J I £ ¶ '4 òÑ