求微分方程y' -y sinx =-sinxcosx的通解 求数学大拿帮忙。。谢谢谢谢谢~！~~~~ 求解微分方程:y'-y=-sinx
\u6c42\u5fae\u5206\u65b9\u7a0by''' =sinx-cosx\u7684\u901a\u89e3\u5f80\u56de\u63a8\u5c31\u884c\u4e86\uff0cy''=-cosx-sinx+c,y'=-sinx+cosx+c,y=cosx+sinx+c(c\u4e3a\u5e38\u6570)
\u5177\u4f53\u56de\u7b54\u5982\u4e0b\uff1a
\u5148\u6c42\u9f50\u6b21\u7ebf\u6027\u5fae\u5206\u65b9\u7a0b\uff1ady/dx=y
lny=c+x
y=e^(x+c)
\u5e38\u6570\u53d8\u5f02
y=c(x)e^x
dy/dx=dc(x)/dx*e^x+c(x)*e^x
\u5e26\u5165\u539f\u65b9\u7a0b\u5f97\uff1adc(x)/dx=-sin(x)*e^(-x)
\u4e24\u8fb9\u540c\u65f6\u79ef\u5206\u5f97\uff1ac(x)=-1/2(sin(x)+cos(x))*e^(-x)+c
\u5e26\u5165\uff1ay=-1/2(sin(x)+cos(x))+c*e^x
\u7ea6\u675f\u6761\u4ef6\uff1a
\u5fae\u5206\u65b9\u7a0b\u7684\u7ea6\u675f\u6761\u4ef6\u662f\u6307\u5176\u89e3\u9700\u7b26\u5408\u7684\u6761\u4ef6\uff0c\u4f9d\u5e38\u5fae\u5206\u65b9\u7a0b\u53ca\u504f\u5fae\u5206\u65b9\u7a0b\u7684\u4e0d\u540c\uff0c\u6709\u4e0d\u540c\u7684\u7ea6\u675f\u6761\u4ef6\u3002
\u5e38\u5fae\u5206\u65b9\u7a0b\u5e38\u89c1\u7684\u7ea6\u675f\u6761\u4ef6\u662f\u51fd\u6570\u5728\u7279\u5b9a\u70b9\u7684\u503c\uff0c\u82e5\u662f\u9ad8\u9636\u7684\u5fae\u5206\u65b9\u7a0b\uff0c\u4f1a\u52a0\u4e0a\u5176\u5404\u9636\u5bfc\u6570\u7684\u503c\uff0c\u6709\u8fd9\u7c7b\u7ea6\u675f\u6761\u4ef6\u7684\u5e38\u5fae\u5206\u65b9\u7a0b\u79f0\u4e3a\u521d\u503c\u95ee\u9898\u3002
这可以用积分因子对于微分方程 y'+P(x)y=Q(x),积分因子为exp(∫P(x)dx)
两边同乘exp(∫-sinx dx)=exp(cosx)
exp(cosx)y' +y exp(cosx)(-sinx) =-sinxcosx*exp(cosx)
(exp(cosx)y)'=-sinxcosx*exp(cosx)
d(exp(cosx)y)/dx=-sinxcosx*exp(cosx)
d(exp(cosx)y)=-sinxcosx*exp(cosx)dx
两边积分
exp(cosx)y=∫-sinxcosx*exp(cosx)dx
=∫cosx*exp(cosx)d(cosx)
因为∫t exp(t) dt =∫ t d(exp(t))=texp(t)-∫exp(t)dt=(t-1)exp(t)
exp(cosx)y=(cosx-1)exp(cosx)+C
y= cosx-1 +C*exp(-cosx)
C是待定常数
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