高数:求下列微分方程的通解(要有过程)
1(e(x+y)-e(x))dx+(e(x+y)+e(x))dy=02(y+1)(2)dy/dx+x(3)=03(x(3)+y(3))dx-3xy(2)dy=0...
1 (e(x+y)-e(x))dx+(e(x+y)+e(x))dy=0
2 (y+1)(2)dy/dx+x(3)=0
3 (x(3)+y(3))dx-3xy(2)dy=0 展开
2 (y+1)(2)dy/dx+x(3)=0
3 (x(3)+y(3))dx-3xy(2)dy=0 展开
2个回答
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解:1.∵(e^(x+y)-e^x)dx+(e^(x+y)+e^x)dy=0
==>(e^y-1)dx+(e^y+1)dy=0
==>(e^y+1)/(e^y-1)dy+dx=0
==>dy+dx=2/(e^y-1)dy
==>y+x=2∫dy/(e^y-1) (等式两边取积分)
==>y+x=2∫e^(-y)dy/(1-e^(-y))
==>y+x=2∫d(1-e^(-y))/(1-e^(-y))
==>y+x+C1=2ln|1-e^(-y)| (C1是积分常数)
==>e^(x+y+C1)=(1-e^(-y))²
==>(1-e^(-y))²=e^C1*e^(x+y)
==>(1-e^(-y))²=Ce^(x+y) (令C=e^C1)
∴原方程的解是:(1-e^(-y))²=Ce^(x+y) (C是积分常数)
2.∵(y+1)²dy/dx+x³=0
==>(y+1)²dy+x³dx=0
==>(y+1)²d(y+1)+x³dx=0
==>(y+1)³/3+x^4/4=C/12 (C是积分常数)
==>4(y+1)³+3x^4=C/12
∴原方程的解是:4(y+1)³+3x^4=C/12 (C是积分常数)
3.∵(x³+y³)dx-3xy²dy=0
==>x³dx+y³dx-3xy²dy=0
==>x³dx+y³dx-xd(y³)=0
==>x²dx=(xd(y³)-y³dx)/x²
==>d(x³)/3=d(y³/x)
==>x³/3+C/3=y³/x (C是积分常数)
==>x(x³+C)=y³
==>y³=x^4+Cx
∴原方程的解是:y³=x^4+Cx (C是积分常数)。
==>(e^y-1)dx+(e^y+1)dy=0
==>(e^y+1)/(e^y-1)dy+dx=0
==>dy+dx=2/(e^y-1)dy
==>y+x=2∫dy/(e^y-1) (等式两边取积分)
==>y+x=2∫e^(-y)dy/(1-e^(-y))
==>y+x=2∫d(1-e^(-y))/(1-e^(-y))
==>y+x+C1=2ln|1-e^(-y)| (C1是积分常数)
==>e^(x+y+C1)=(1-e^(-y))²
==>(1-e^(-y))²=e^C1*e^(x+y)
==>(1-e^(-y))²=Ce^(x+y) (令C=e^C1)
∴原方程的解是:(1-e^(-y))²=Ce^(x+y) (C是积分常数)
2.∵(y+1)²dy/dx+x³=0
==>(y+1)²dy+x³dx=0
==>(y+1)²d(y+1)+x³dx=0
==>(y+1)³/3+x^4/4=C/12 (C是积分常数)
==>4(y+1)³+3x^4=C/12
∴原方程的解是:4(y+1)³+3x^4=C/12 (C是积分常数)
3.∵(x³+y³)dx-3xy²dy=0
==>x³dx+y³dx-3xy²dy=0
==>x³dx+y³dx-xd(y³)=0
==>x²dx=(xd(y³)-y³dx)/x²
==>d(x³)/3=d(y³/x)
==>x³/3+C/3=y³/x (C是积分常数)
==>x(x³+C)=y³
==>y³=x^4+Cx
∴原方程的解是:y³=x^4+Cx (C是积分常数)。
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