请教微分方程:y'=1/(x+siny)

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请教微分方程:y'=1/(x+siny)
请教微分方程:y'=1/(x+siny)

请教微分方程:y'=1/(x+siny)
dx/dy=x+siny
x=e^(∫dy)(∫sinye^(∫-dy)dy+C)
=e^y*[-1/2*e^(-y)*(siny+cosy)+C]
=Ce^y-1/2*(siny+cosy)是原方程的解
C是任意常数

y'=1/(x+siny)
dx/dy=x+siny
dx/dy=x
dx/x=dy
dlnx=dy
y=lnx+C
x=C'e^y
x=C(y)e^y
C'(y)e^y=siny
C'(y)=siny*e^(-y)
C(y)=∫sinye^(-y)dy=-∫e^(-y)dcosy=-e^(-y)cosy-∫cosye...

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y'=1/(x+siny)
dx/dy=x+siny
dx/dy=x
dx/x=dy
dlnx=dy
y=lnx+C
x=C'e^y
x=C(y)e^y
C'(y)e^y=siny
C'(y)=siny*e^(-y)
C(y)=∫sinye^(-y)dy=-∫e^(-y)dcosy=-e^(-y)cosy-∫cosye^(-y)dy
=-e^(-y)cosy-∫e^(-y)dsiny=-e^(-y)cosy-e^(-y)siny-∫sinye^(-y)dy
∫sinye^(-y)= -e^(-y)(cosy+siny)/2+C
C(y)=-e^(-y)(cosy+siny)/2+C
x=-(cosy+siny)/2+Ce^(y)

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