DE - Solutions By Substitutions (algebraic manipulation problem)

[Nano-Passion]

I can't for the life of me understand how the book went from step A to step B--even though I realize its just a pesky algebraic manipulation.

Step A

$$M(1,u) dx +N(1,u)[u dx+ x du] = 0$$

Step B

$$[M(1,u) + uN(1,u) ]dx + xN(1,u)du = 0$$

Step A to Step B? I tried factoring and tinkering around here and there but all was in vain.

 

[SammyS]

I can't for the life of me understand how the book went from step A to step B--even though I realize its just a pesky algebraic manipulation.

Step A

$$M(1,u) dx +N(1,u)[u dx+ x du] = 0$$

Step B

$$[M(1,u) + uN(1,u) ]dx + xN(1,u)du = 0$$

Step A to Step B? I tried factoring and tinkering around here and there but all was in vain.

You'll be embarrassed if we work it out for you !

$M(1,u) dx +N(1,u)[u dx+ x du] = 0$ → $M(1,u) dx + u N(1,u) dx+ x N(1,u) du = 0$

 

[Nano-Passion]

You'll be embarrassed if we work it out for you !

$M(1,u) dx +N(1,u)[u dx+ x du] = 0$ → $M(1,u) dx + u N(1,u) dx+ x N(1,u) du = 0$

To me it just seems that you wrote step B in an extremely similar form. I don't understand how you got to that step either. There must be something I'm misinterpreting because it seems like your missing at least a couple terms (if you factor).

 

[SammyS]

To me it just seems that you wrote step B in an extremely similar form. I don't understand how you got to that step either. There must be something I'm misinterpreting because it seems like your missing at least a couple terms (if you factor).

Distribute the N(1,u):

$N(1,u)[u dx+ x du]$​

 

[Nano-Passion]

Distribute the N(1,u):

$N(1,u)[u dx+ x du]$​

I always thought one must distribute everything? Given how we have two polynomials, shouldn't you always foil? Take for example:

$$(x+2)(x-2)$$ It wouldn't make sense to only distribute the 2 in the first bracket and say the expression is equal to
$$x+2x-4$$

 

[SammyS]

I always thought one must distribute everything? Given how we have two polynomials, shouldn't you always foil? Take for example:

$$(x+2)(x-2)$$ It wouldn't make sense to only distribute the 2 in the first bracket and say the expression is equal to
$$x+2x-4$$

Actually, that's $x^2+2x-4$

but for your problem ...

a + b(c+e) is a + bc + be . --- Right ?

The a is M(1,u)dx ,

the b is N(1,u) ,

the c is u dx ,

and the e is x du

 

[Nano-Passion]

Actually, that's $x^2+2x-4$

I know it is x^2+2x-4. But I was saying that by your method it would have been x+2x-4.

a + b(c+e) is a + bc + be . --- Right ?

The a is M(1,u)dx ,

the b is N(1,u) ,

the c is u dx ,

and the e is x du

Your logic is spot on. But something is still bothering me.

a + b(c+e) is a + bc + be

The a is x
The b is +2
The c is x
The e is -2

So then it would follow
$$a+bc+be$$
$$x+2x-4$$

So What happened here? It doesn't come out to x^2+2x-4 like I was trying to say before.

The correct answer would have been ac+bc+be. I know what you said is consistent but something is being mixed up here.

Edit:

Maybe what makes the difference is that there aren't brackets aroudn the first two terms. So that

$$(a+b) (c+e) ≠ a+b(c+e)$$

Is this correct?

 

[SammyS]

I always thought one must distribute everything? Given how we have two polynomials, shouldn't you always foil? Take for example:

$$(x+2)(x-2)$$ It wouldn't make sense to only distribute the 2 in the first bracket and say the expression is equal to
$$x+2x-4$$

I misread what you said here.

Yes, of course $(x+2)(x-2)$ is the same as $x^2+2x-4\ .$

What you have in this problem is more like $x+2(x-2)\,,$ which is indeed the same as $x+2x-4\ .$

In Step A) of your original post, the M(1,u) does not get distributed through anything.

 

[Bohrok]

Maybe what makes the difference is that there aren't brackets aroudn the first two terms. So that

$$(a+b) (c+e) ≠ a+b(c+e)$$

Is this correct?

That's correct. Parentheses can make all the difference in an answer.

 

[Nano-Passion]

I misread what you said here.

Yes, of course $(x+2)(x-2)$ is the same as $x^2+2x-4\ .$

What you have in this problem is more like $x+2(x-2)\,,$ which is indeed the same as $x+2x-4\ .$

In Step A) of your original post, the M(1,u) does not get distributed through anything.

Oh thanks!

That's correct. Parentheses can make all the difference in an answer.

Wow, that is a really important algebraic property to know. I find it funny that I learn it now while in differential equations! :rofl: