13.æ±ç±æ¹ç¨x/z=ln(z/y)æç¡®å®çå½æ°z=f(xï¼y)çå
¨å¾®åã
解ï¼ä½å½æ°F(xï¼yï¼z)=x/z-ln(z/y)=0
å∂z/∂x=-(∂F/∂x)/(∂F/∂z)=-(1/z)/[-(x/z²)-(1/y)/(z/y)]=z/(x+z)
∂z/∂y=-(∂F/∂y)/(∂F/∂z)=-[(-z/y²)/(z/y)]/[-(x/z²)-(1/y)/(z/y)]
=-z²/[y(x+z)]
æ
dz=[z/(x+z)]dx-[z²/y(x+z)]dy
19.æ±ç±æ¹ç¨x+2y+z-2â(xyz)=0æç¡®å®çå½æ°z=f(xï¼y)çå
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解ï¼è®¾F(xï¼yï¼z)=x+2y+z-2â(xyz)=0,åï¼
∂z/∂x=-(∂F/∂x)/(∂F/∂z)=-[1-(yz/âxyz)]/[1-(xy/âxyz)]
=-[â(xyz)-yz]/[â(xyz)-xy]
∂z/∂y=-(∂F/∂y)/(∂F/∂z)=-[2-(xz/âxyz)]/[[1-(xy/âxyz)]
=-[2â(xyz)-xz]/[â(xyz)-xy]
æ
dz={-[â(xyz)-yz]dx-[2â(xyz)-xz]dy}/[â(xyz)-xy]
20.æ±å¾®åæ¹ç¨y'+x/y=0满足åå§æ¡ä»¶y(0)=4çç¹è§£
解ï¼dy/dx=-x/yï¼å离åéå¾ydy=-xdxï¼
积åä¹å¾é解为 y²/2=-x²/2+Cï¼
代å
¥åå§æ¡ä»¶y(0)=4å¾C=8ï¼æ
å¾ç¹è§£ä¸º y²=-x²+16
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