First order linear ODE problem

[Marian]

Hello, everyone,

this is my first time posting something here. I hope this is not too simple.

I'm studying the book: Schawm's Easy Outlines in Differential Equations [link].

I'm trying to understand a solved problem. This is: page 18, problem 2.7
Probably it is just a notation problem, but I'm not sure.

My question is what happens with the second term (4yx^3) when the expression is written in Leibniz form:


Something similar happens in the problem 2.8

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The complete problems:

Problem 2.7

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Problem 2.8

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Thank you very much. If there is anything unclear, please tell me.

Best,
Marian

 

[HallsofIvy]

I would say that you need to review Calculus. By the "product rule", (uv)'= u'v+ uv'. In problem 2.7, \(\displaystyle u= x^4\) and \(\displaystyle v= y\)

\(\displaystyle (x^4y)'= (x^4)'u+ (x^4)u'= 4x^3u+ x^4u'\).

Same thing with problem 2.8: \(\displaystyle (ze^{-x^2/2})'= z'(e^{-x^2/2}+ z(e^{-x^2/2})'= e^{-x^2/2}z'- 2xe^{-x^2/2}z\).

 

[Marian]

Use of integrating factors for solvinf linear ODEs

I would say that you need to review Calculus. By the "product rule", (uv)'= u'v+ uv'. In problem 2.7, \(\displaystyle u= x^4\) and \(\displaystyle v= y\)

\(\displaystyle (x^4y)'= (x^4)'u+ (x^4)u'= 4x^3u+ x^4u'\).

Same thing with problem 2.8: \(\displaystyle (ze^{-x^2/2})'= z'(e^{-x^2/2}+ z(e^{-x^2/2})'= e^{-x^2/2}z'- 2xe^{-x^2/2}z\).

I did not explain exactly what "what happens with the term" meant.
In the same images, the first expression is accompanied by another:

expression 1 or expression 2

I did not get how expression 1 could become expression 2.

In corresponding theory for those problems (use of integrating factors to solve linear ODE),
this formula is used to transform from expression 1 to expression 2.
Note that expression 1 has been multiplied by a specific integrating factor I(x).

The formula is:





I'm really sorry for not being patient enough and just posting my question.
Now I got to understand it myself.
I have been trying to understand during this week, and after posting here, it just clicked in my mind.

Thank you a lot for your attention!

 

[Steven G]

I would say that you need to review Calculus. By the "product rule", (uv)'= u'v+ uv'. In problem 2.7, \(\displaystyle u= x^4\) and \(\displaystyle v= y\)

\(\displaystyle (x^4y)'= (x^4)'u+ (x^4)u'= 4x^3u+ x^4u'\).

Same thing with problem 2.8: \(\displaystyle (ze^{-x^2/2})'= z'(e^{-x^2/2}+ z(e^{-x^2/2})'= e^{-x^2/2}z'- 2xe^{-x^2/2}z\).

Marian, HallsofIvy is completely correct in the above response but unless I am missing something I think that the u and u' should be v and v'.

 

[HallsofIvy]

I would say that you need to review Calculus. By the "product rule", (uv)'= u'v+ uv'. In problem 2.7, \(\displaystyle u= x^4\) and \(\displaystyle v= y\)

\(\displaystyle (x^4y)'= (x^4)'u+ (x^4)u'= 4x^3u+ x^4u'\).

This should, of course, have been \(\displaystyle (x^4y)'= (x^4)'y+ (x^4)y'= 4x^3y+ x^4y'\).

Same thing with problem 2.8: \(\displaystyle (ze^{-x^2/2})'= z'(e^{-x^2/2}+ z(e^{-x^2/2})'= e^{-x^2/2}z'- 2xe^{-x^2/2}z\).

 

[Marian]

This should, of course, have been \(\displaystyle (x^4y)'= (x^4)'y+ (x^4)y'= 4x^3y+ x^4y'\).

Thank you for the correction.

I'm not saying is in any way wrong. It is just a different thing what I needed to know. Not the developement of the differential of that term.
I was confused on why that term "dissapeared" in the seccond expression
(remember the first images? expression 1 or espression 2).

It does not "dissapear", because expression 2 is not the derivative of expression 1, but it is the substitution of the formula I posted in my second answer (use of integral factors)

Thank you very much for your answers!