Question #138694
Using X=x-2, Y=y+1, reduce the equation
4(x-2)^2.dy/dx=(x+y-1)^2 to the homengenous form of 1st order equation
1
Expert's answer
2020-10-19T16:34:13-0400

A function f(x,y) is said to be homogeneous of degree n if the equation f(zx,zy)=znf(x,y)f(zx,zy)=z^nf(x,y) .

A first‐order differential equation M(x,y)dx+N(x,y)dy=0M(x,y)dx+N(x,y)dy=0 is said to be homogeneous if M(x,y)M(x,y) and N(x,y)N(x,y) are both homogeneous functions of the same degree.


4(x2)2dydx=(x+y1)24(x-2)^2\frac{dy}{dx}=(x+y-1)^2 ;

if X=x2X=x-2 , Y=y+1Y=y+1, dYdX=d(y+1)d(x2)=dydx\frac{dY}{dX}=\frac{d(y+1)}{d(x-2)}=\frac{dy}{dx} than:

4(X)2dYdX=((x2)+(y+1))24(X)^2\frac{dY}{dX}=((x-2)+(y+1))^2 ;

4X2dYdX=(X+Y)24X^2\frac{dY}{dX}=(X+Y)^2 ;

(X+Y)2dX4X2dY=0(X+Y)^2dX-4X^2dY=0 .

where

M(X,Y)=(X+Y)2M(X,Y)=(X+Y)^2 and N(X,Y)=4X2N(X,Y)=-4X^2 are homogeneous functions of the same degree (namely, 2), because:

M(zX,zY)=(zX+zY)2=z2(X+Y)2M(zX,zY)=(zX+zY)^2=z^2(X+Y)^2 and

N(zX,zY)=4(zX)2=z2(4X2)N(zX,zY)=-4(zX)^2=z^2(-4X^2) .

Answer: the reduced equation is (X+Y)2dX4X2dY=0(X+Y)^2dX-4X^2dY=0 .


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