微分方程y'=(x+y+3)/(x+2y-1)的解

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微分方程y'=(x+y+3)/(x+2y-1)的解
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微分方程y'=(x+y+3)/(x+2y-1)的解
微分方程y'=(x+y+3)/(x+2y-1)的解

微分方程y'=(x+y+3)/(x+2y-1)的解
令X=x+7,Y=y-4,
x+y+3=(x+7)+(y-4)=X+Y
x+2y-1=(x+7)+2(y-4)=X+2Y
则dy/dx=dY/dX
原微分方程化为:
dY/dX=(X+Y)/(X+2Y) 齐次方程
分子分母同除以X得:dY/dX=(1+Y/X)/(1+2Y/X)
令u=Y/X,得Y=uX,dY/dX=u+X*du/dX
上式化为:u+X*du/dX=(1+u)/(1+2u)
du/dX=(1+u)/(1+2u)-u=(1-2u^2)/(1+2u)
则(1+2u)/(1-2u^2)du=dX
得:1/(1-2u^2)du+2u/(1-2u^2)du=dX
积分得:√2/4*ln|(2u+√2)/(2u-√2)|-(1/2)ln|1-2u^2|=X+C
将u换成Y/X得:√2/4*ln|(2Y+√2X)/(2Y-√2X)|-(1/2)ln|1-2(Y/X)^2|=X+C
将X换成x+7,Y换成y-4得,两边再乘以2:解为
√2/2*ln|(2(y-4)+√2(x+7))/(2(y-4)-√2(x+7))|-ln|1-2((y-4)/(x+7))^2|=2x+C1