MHB Solving Substitution Steps: Examples Explained

[find_the_fun]

In the context of "solutions by substitutions" in the examples there's a step I don't understand:

If we let $$y = ux$$ then $$dy = u dx + x du$$
-------
we have $$u = y^{-1}$$ or $$y = u^{-1}$$. We then substitute $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} = -u^{-2} \frac{du}{dx}$$

What is going on in these two examples? If y=ux then [math]\frac{dy}{dx}=u[/math] and [math]dy=u dx \neq udx + xdu[/math]

Same for the other, if [math]y=u^{-1}[/math] then [math]\frac{dy}{dx}=0[/math]?

 

[Prove It]

In the context of "solutions by substitutions" in the examples there's a step I don't understand:
What is going on in these two examples? If y=ux then [math]\frac{dy}{dx}=u[/math] and [math]dy=u dx \neq udx + xdu[/math]

Same for the other, if [math]y=u^{-1}[/math] then [math]\frac{dy}{dx}=0[/math]?

u is a function of x, so to differentiate ux you have to use the product rule, giving (ux)' = u + u'x

 

[find_the_fun]

But how do you know u is a function of x?

 

[MarkFL]

But how do you know u is a function of x?

Restricting $u$ to be a constant would not allow $y$ to be a general function of $x$.

 

[Prove It]

More specifically, your lecturer has explained the steps in a roundabout way. What you really need to do is to get your DE to be a function of (y/x), so that you can substitute u = y/x.

Notice that u will only be a constant if y = x (which it most likely won't).

Thus in general u must be a function of x.