Why can one do this "trick" in the First Order Linear D.E. method?

If we have a first order d.e. like: $$\frac {dy} {dx} - \frac y x = 1$$
I would use two subs, namely: ##y = uv## and ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx}##

So I get: ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx} - \frac {uv} {x} = 1##

I then factor like this: ##\frac {dy} {dx} = u \frac {dv} {dx} + v (\frac {du} {dx} - \frac {u} {x}) = 1##

I am then told that I can make ##\frac {du} {dx} - \frac {u} {x} = 0##. *Why* can I do this?

Ebby said:

Homework Statement: Explain the First Order Linear D.E. method
Relevant Equations: Can you explain the step shown below?

If we have a first order d.e. like: $$\frac {dy} {dx} - \frac y x = 1$$
I would use two subs, namely: ##y = uv## and ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx}##

So I get: ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx} - \frac {uv} {x} = 1##

I then factor like this: ##\frac {dy} {dx} = u \frac {dv} {dx} + v (\frac {du} {dx} - \frac {u} {x}) = 1##

I am then told that I can make ##\frac {du} {dx} - \frac {u} {x} = 0##. *Why* can I do this?

Are you sure the trick isn't ## y = u x ##?

$$ \frac{dy}{dx} = \frac{du}{dx}x + u $$

Then sub for ## \frac{dy}{dx}##

$$\frac{du}{dx}x + u - u = 1 $$

$$ \implies \frac{du}{dx} = \frac{1}{x} $$

?

Just solve ##u’(x) = u/x##. This is a simple homogeneous ODE with the general solution ##u(x) = Ax## for any constant ##A##.

If the question is why you are allowed to pick a ##u## such that the ODE is satisfied: The substitution is valid for any ##u##.* The trick is to pick a ##u## such that the resulting ODE for ##v## takes a simpler form that allows you to solve it.

* Not any function of course. Suitable requirements on differentiability and not introducing additional zeros apply.

Ebby said:

Homework Statement: Explain the First Order Linear D.E. method
Relevant Equations: Can you explain the step shown below?

If we have a first order d.e. like: $$\frac {dy} {dx} - \frac y x = 1$$
I would use two subs, namely: ##y = uv## and ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx}##

So I get: ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx} - \frac {uv} {x} = 1##

I then factor like this: ##\frac {dy} {dx} = u \frac {dv} {dx} + v (\frac {du} {dx} - \frac {u} {x}) = 1##

I am then told that I can make ##\frac {du} {dx} - \frac {u} {x} = 0##. *Why* can I do this?

I doubt it too, just like @erobz . If you could do that, I your second expression in the equality would be 0, and in your chain of equalities, you would end up with ##\frac {dy}{dx}=1##, from which it follows that ##f(x)=x## is a solution, which it isn't , for your initial equation.

Ebby said:

Homework Statement: Explain the First Order Linear D.E. method
Relevant Equations: Can you explain the step shown below?

*Why* can I do this?

You can choose any ##u## OR any ##v## that makes the solution process easier. If you pick a certain ##u##, you got to adjust the ##v## accordingly so that ##v=\frac{y}{u}## and similar if you chose a certain ##v##.

Beware that the "OR" is exclusive, that is you are not free to choose both of ##u ,v## because as i said before if you choose one the other becomes chosen already as y/"what you chose at first step".

Try choosing as ##v=c## that is a constant and see what happens too.

he has a typo in his second long equality

It should be $$\frac{dy}{dx}-\frac{y}{x}=...$$

WWGD said:

I doubt it too, just like @erobz . If you could do that, I your second expression in the equality would be 0, and in your chain of equalities, you would end up with ##\frac {dy}{dx}=1##, from which it follows that ##f(x)=x## is a solution, which it isn't , for your initial equation.

As @Delta2 says, that’s just a typo in the OP. The original statement is

Ebby said:

##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx}##

and the typo comes right after:

Ebby said:

So I get: ##\frac {dy} {dx} = u \frac {dv} {dx} + v \frac {du} {dx} - \frac {uv} {x} = 1##

where OP has forgotten to add ##-y/x## to the RHS.

The actual ODE leads to
$$
\frac{dv}{dx} = \frac{1}{Ax}
$$
where ##y(x) = Ax \,v(x)##.

I don't know what this whole idea about making substitutions is but I'm sure it amounts to the following:

The idea is to make the left side into a product rule via an integrating factor

If you have the equation

##\frac{dy}{dx} +p(x) y = g(x)##

If you multiply through by an "integrating factor" ## \mu \left(x\right) = e^{\int p(x)\,dx}## you will make the left hand side into a product rule.

In your case

##\frac{dy}{dx} + \left( - \frac{1}{x} \right) y = 1##

multiply through by ##\mu \left(x \right) = e^{\int -\frac{1}{x} \,dx} = \frac{1}{x}## which leads to

##\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x}##

Recognize the left as a product rule

##\frac{d}{dx} \left( \frac{y}{x}\right) = \frac{1}{x}##

Integrate both sides accounting for a "constant of integration" then isolate ##y## from ##x##

Plug it into the original differential equation and it works

This strategy is pretty standard, more on the "integrating factor" https://tutorial.math.lamar.edu/classes/de/linear.aspx

PhDeezNutz said:

I don't know what this whole idea about making substitutions is but I'm sure it amounts to the following:

The idea is to make the left side into a product rule via an integrating factor

If you have the equation

##\frac{dy}{dx} +p(x) y = g(x)##

If you multiply through by an "integrating factor" ## \mu \left(x\right) = e^{\int p(x)\,dx}## you will make the left hand side into a product rule.

In your case

##\frac{dy}{dx} + \left( - \frac{1}{x} \right) y = 1##

multiply through by ##\mu \left(x \right) = e^{\int -\frac{1}{x} \,dx} = \frac{1}{x}## which leads to

##\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x}##

Recognize the left as a product rule

##\frac{d}{dx} \left( \frac{y}{x}\right) = \frac{1}{x}##

Integrate both sides accounting for a "constant of integration" then isolate ##y## from ##x##

Plug it into the original differential equation and it works

This strategy is pretty standard, more on the "integrating factor" https://tutorial.math.lamar.edu/classes/de/linear.aspx

This is similar to the idea for finding weak solutions to DEs.

WWGD said:

This is similar to the idea for finding weak solutions to DEs.

Didn't know that but now with that knowledge I have a direction to look in with insight you have given me!

I'm a bit confused because apparently I've made a typo but I simply used an example from this site about DEs (it's Example 1): https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html

Where exactly did I go wrong?

Ebby said:

I'm a bit confused because apparently I've made a typo but I simply used an example from this site about DEs (it's Example 1): https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html

Where exactly did I go wrong?

Ehm, except the typo you didn't do anything else wrong, you are right that you can choose a ##u## such that
$$\frac{du}{dx}-\frac{u}{x}=0$$, or any other condition that helps solve the ODE.

OK, I see the typo now. Phew. But let me ask the really dumb question:

Given that: $$u \frac {dv} {dx} + v(\frac {du} {dx} - \frac u x) = 1$$

Why can we say that, algebraically: $$\frac {du} {dx} - \frac u x = 0$$

? I know, I'm really stupid...

You got to try to read -while paying some attention -of all the previous posts here

But anyway I am gonna say it again.

You CAN CHOOSE a ##u## such that ##\frac{du}{dx}-\frac{u}{x}=0##. You are free to choose any ##u## you like. You are also free to choose any ##v## you like. But not both.

Ebby said:

OK, I see the typo now. Phew. But let me ask the really dumb question:

Given that: $$u \frac {dv} {dx} + v(\frac {du} {dx} - \frac u x) = 1$$

Why can we say that, algebraically: $$\frac {du} {dx} - \frac u x = 0$$

? I know, I'm really stupid...

Post #4, extra emphasis:

Orodruin said:

If the question is why you are allowed to pick a ##u## such that the ODE is satisfied: The substitution is valid for any ##u##.* The trick is to pick a ##u## such that the resulting ODE for ##v## takes a simpler form that allows you to solve it.

* Not any function of course. Suitable requirements on differentiability and not introducing additional zeros apply.

Ok I kinda am within your mind and now you ask
"Why I am free to choose any u i like???, u, v are supposed to be specific functions"

##u , v## are just functions that satisfy ##y=uv## (and y satisfies that given ODE).

But ##u,v## are not so specific.
Lets say for example that ##y=x^4+x^2## (i dont think that this y satisfies the given ODE but it doesnt matter you ll see my point as you go on reading here)

You can write ##y=x^2(x^2+1)## or ##y=x^3(x+\frac{1}{x})##

So you can choose as ##u=x^2, v=x^2+1## or ##u=x^3, v=x+\frac{1}{x}##

I think now you should be able to understand why u,v are not so specific.

I mean, you could choose ##u = e^{x + 4i}- \ln(x^x)##, but that would not make the differential equation for ##v## any simpler. The trick is to pick a ##u## that makes the ODE for ##v## simpler. In this case, such a ##u## satisfies that particular ODE because that means a term disappears.

Even if one wants to choose as ##u## as given at post #18, you can choose as ##v=\frac{x^4+x^2}{e^{x+4i}-\ln{x^x}}## and again you will have that ##y=uv## holds.

Thank you. I now understand.

I think if someone had said, "Think that you are choosing to make ## \frac {du} {dx} = \frac u x ## so that the ## v ## term disappears, rather than thinking ## \frac {du} {dx} - \frac u x = 0 ##," it would have been clearer to me. Anyway, thank you all. This has been helpful in a more far reaching sense.