Change of variables in dy/dx = (y-x)^2 + 1 gives v^2 + 1, not v^2

[JasonJo]

Am I insane or is this a typo:

Consider the nonlinear DE

dy/dx = (y-x)^2 + 1

Show that the change of variables, u=x, v = y-x transforms this DE into the seperable DE: dv/du = v^2

dv = dy
du=dx
dv/du = dy/dx = (y-x)^2 + 1 = v^2 + 1

not v^2

am i wrong ?

 

[stunner5000pt]

In your expression for dv, where did dx go??
it should be dv = dy - dx


or you could have just differentiated v wrt x
taht is dv/dx which is the same as dv/du because of the way u is defined.

 

[JasonJo]

ok so dv/dx = -1, so dv/du = -1. so dv/du = -1(dy/dx), which is:

-1 - (y-x)^2 or -1 - v^2

i got that dv/du = dy/dx = (y-x)^2 + 1, but i can't get that dv/du = v^2 by itself.

 

[stunner5000pt]

ok so dv/dx = -1 , so dv/du = -1. so dv/du = -1(dy/dx), which is:

-1 - (y-x)^2 or -1 - v^2

i got that dv/du = dy/dx = (y-x)^2 + 1, but i can't get that dv/du = v^2 by itself.

ok this time you went and eliminated the y

this is what it should be
$$\frac{dv}{du} = \frac{dv}{dx} = \frac{dy}{dx} - \frac{dx}{dx} = \frac{dy}{dx} - 1$$
v and y are both funcions of x, so when you differentiate eitehr you get dv/dx and dy/dx.

 

[arildno]

I can't see the point of doing the u-substitution.

Just define $$v(x)=y(x)-x\to{y}(x)=v(x)+x\to\frac{dy}{dx}=\frac{dv}{dx}+1$$

A simple substitution then yields:
$$\frac{dv}{dx}=v^{2}$$

 

[JasonJo]

ahhhh!

thank you guys so much!