Solving a Differential Equation by separating the variables (2) : ExamSolu

Differential equation separating the variables.

[ExamSolutions_Maths]

It could be as there can be several different forms of the answer where constants are concerned.

[munansangu]

i love the way to make maths easy! thanks

[ahmedsyed]

The display of TWO methods is fascinating. Thank you.

[rosemarymchangila]

I have never seen me a better maths teacher! Keep it up!

[RShahProductions]

You divided by y and (x-4); I divided by 3y and (x-4) => 1/(x-4) = 1/3y (dy/dx)
Integrating from there brought me the exact same result as you except my +C was on the other side, meaning my ln division was the inverse of yours.
Does it matter?

[ab]

here, if take log |c| in left side than after multiply by y we get y equals to (x-4)^3/A which is different from ans. in this video. In pervious video you proves that it doesn't matter that at which side we take constant. So, plz. reply to correct my mistake it helps me in my studies.

[jeffhart]

This was a very helpful tutorial. Thanks very much! :-)

[garykavanagh]

These videos are a great help, Thanks for uploading,

[gendrygaming]

what if I leave 3 with y before I integrate it? is it okay?

[mathsgreek]

excellent ! antiloging means to equal the log s??

[humairaahmed]

Hi! Could you please tell me why integrating 1/y equals ln mod y? Why not without mod? Or could u give me the link to video explaining this, if there is a video on it? (BTW, I have just learnt the derivative of ln functions using your website) .

[lanature]

why do we have to send the 3 over (x-4) and why cant i just integrate it as 3y?

[ExamSolutions_Maths]

That is okay. Give me your answer and I will show you how it converts to mine.

[comeonuglos]

Hi,

like Richard I divided by 3y instead of just y so I got a different answer to yours and I was wondering whether you could tell me if it is correct.

My answer
y='cubed root of'(A(x-4))

Cheers

[zsylechano]

hi good day.. i'll just wanna ask if the second approach is the example for not making "y" as the subject? Are there any other approaches, or it's just the two ways that you have shown?.. thank you. :)

[Den-blss]

how do i know if its the particular solution and not the general solution.. how do i know if i should stop solving? sorry for bad english.

[ramborai]

you are just awesome...at the end you said any help would be appreciated....what kind of help do  you need???

[brandongregory]

why not just multiply through by e, to cancel the Logs? I did this method and got a different answer :/

[georgiadixson]

i'm sorry if this is a stupid question, but why is the integral of 1/y log of the modulus of y??

[kittykatmai]

is this okay as well:
ln|(x-4)^3| = ln|y| + C
ln|(x-4)^3| - ln|y| = C
ln|(x-4)^3/y|= C