(y"+tanxy'-tanxy'+y)=(x²+1+cosx)(cosx)y"-(sinx)y'+(sinx)y'+(cosx)y=(x²+1+cosx)cosx(cosx)y'+(sinx)y=∫x²cosx+cosx+cos²x dxx²cosx=x²cosx+2xsinx-2xsinx+2cosx-2cosx∫x²c...谢谢~(secx)y'+(tanxsecx)y=x²sinx/cos²x+2x/cosx-sinx/cos²x+sinx/(2cosx)+x/(2cos²x)+C/cos²x
(secx)y=∫{x²sinx/cos²x+2x/cosx)-(sinx/cos²x+cosx(tanx))+sinx+(0.5tanx
+0.5xsec²x)+Csec²x} dx
我说觉得不对呢--------------------------------------------------------这下一步不对，积分求错了。不知道哪多了个-sinx/cos²x
(secx)y= x²secx -sinx(tanx)-cosx+0.5xtanx+Ctanx+D
y=x²-sinxcosxtanx-cos²x+0.5xtanxcosx+Ctanxcosx+Dcosx
y=x-sin²x-cos²x+0.5xsinx+Csinx+Dcosx
y=x-1+0.5xsinx+Csinx+Dcosx