Check solution to DE

Question

I was solving the following DE

$$\frac{y+x}{x}=y'$$

and my solution was

$$xdy = (y+x)dx$$

Letting $y=xv$ then $dy=xdv+vdx$ and so the above equation becomes

$$x^2dv + xvdx =xvdx+xdx$$

From here I get

$$x^2dv=xdx$$

Now here is where things get interesting. I can write this as $$\int 1 dv = \int \frac{1}{x}dx$$ And get the solution $$y=x\ln(x)+cx$$

But when I took my exam, I didn't do this.... instead I did this:

$$\int x^2dv = \int x dx$$ Which gives

$$x^2v=\frac{1}{2}x^2+C$$ $$xy=\frac{1}{2}x^2+c$$ $$y=\frac{1}{2}x+\frac{c}{x}$$

How can I reconcile these apparently different solutions?

Answer 1

Note that in the exam, you incorrectly wrote that $$ \int x^2\, dv = x^2 v $$ As $v$ depends on $x$, $x$ depends on $v$, so treating $x$ as a constant with respect ot $v$ is wrong.

That the second 'solution' is wrong can also be seen by just checking, we have for $y(x) := \frac 12x + \frac{c}x$ that $$ y'(x) = \frac 12 - \frac c{x^2} $$ and $$ \frac{y(x)+x}{x} = \frac{\frac 32x+\frac cx}x = \frac 32 + \frac c{x^2} \ne y'(x). $$

Answer 2

The mistake is in the second attempt of yours performed at school. It is when you go from this to this:

$$\int x^2dv = \int x dx \implies x^2v=\frac{1}{2}x^2+C$$

This is totally wrong.

Actually, $\int x^2dv = \int x dx \not \implies x^2v=\frac{1}{2}x^2+C$

You must remember $x$ is not a constant, but a function of $v$ related as $x=\frac{y}{v}$. So, that integration of $\int x^2dv$ to $x^2v$ is wrong.

The first procedure is the correct one.

Answer 3

You have stumbled on a subtle point which is usually not explained very well (or at all). Consider a first order differential equation given by $y'(x) = F(x,y)$. Written in "differential" notation, we have

$$ \frac{dy}{dx} = F(x,y). $$

Performing the regular "mindless" manipulations one does when solving ODEs, we can rewrite it as

$$ dy = F(x,y) dx. $$

If your second method would be legal, then we could have just integrated immediately both sides and get

$$ y = \int F(x,y) \, dx + C. $$

Does this mean we have managed to solve all first order equations and render the first few weeks of an ODE course irrelevant? Of course not.

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The question you should ask yourself is not why this method doesn't work but why the other method works. That is, why, when you have an equation of the form $y'(x) = \frac{g(x)}{f(y)}$ and you rewrite it as

$$ f(y) dy = g(x) dx $$

and integrate formally both sides, you actually get an implicit solution $F(y) = G(x) + C$ to the equation. If you formulate it as a theorem and prove it rigorously without introducing the differential bullshit, you'll see that it is actually a non-trivial result that depends on the implicit function theorem and the chain rule. I really recommend you to do this exercise and then return to the "wrong formula"

$$ y = \int F(x,y) \, dx + C $$

and interpret it in a way so that it actually holds (then you'll see it is, unlike in the separable case, completely useless).