Answer to Question #138694 in Differential Equations for Nikhil Singh

Question

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

Answer 1

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

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

$4(x-2)^2\frac{dy}{dx}=(x+y-1)^2$ ;

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

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

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

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

where

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

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

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

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