Question

How did my professor simplify this step: ((200+t)^2) *(dy/dt) + ((200+t)^2) * ((2)/(200+t))y = 18(200+t)^2

to:

d/dt((200+t)^2)*y = 18(200+t)^2

Answer 1

any ideas? i dont know how he gets the left side from the top to the bottom part. I understand the right side of equal sign..

Answer 2

it also says the left side is d/dt (v * y) which i don't understand

Answer 3

Here is how it looks before and after:

Answer 4

@jotopia34 I posted the equation

Answer 5

working it out now

Answer 6

I in no way, shape or form can differentiate and have the (200+t)^2 show up on both sides.....don't know if he has a typo, or I am missing something's sorry!

Answer 7

it doesnt have to show up on both sides, I'm just wondering how the left side in picture 1 gets simplified to the left side in 2nd picture i posted

Answer 8

@jotopia34 ^

Answer 9

Linear differential equation, he most likely used an integration fraction.

Answer 10

You sure want to get it into standard form before you proceed in any way.

Answer 11

@Andysebb, can you do that?

Answer 12

what do you mean? anyway you can show baby steps on how to get the left side in the first pic to the 2nd?

Answer 13

dividing by (200+t)^2 leads to:\[y'+\frac{ 2 }{ 200+t }y=18\]

Now you need to find an integration factor, have you done that before?

Answer 14

an Integration factor would be:\[\mu(t)=\exp ( \int\limits \frac{ 2 }{ 200+t }dt)\]

Answer 15

So an integration factor is:

\[\mu(t)=(200+t)^2\] multiply your original differential equation by that and you will see that the LHS can now be expressed as a product rule of differentiation.

Answer 16

Yes I got the integrating factor before getting to that step. But i dont understand how he simplifies the left side further to be only \[d/dt((200+t)^2)*y \]

Answer 17

well that is the product rule isn't it? The product rule of differentiation. multiply your differential equation by your integration factor.

you should get \[(200+t)^2y'+2(200+t)y=18(200+t)^2\]

and now see that the left hand side is the same as the product rule of differentiation:
\[\frac{ d(y(200+t)^2) }{ dt }=18(200+t)^2\]

Answer 18

does that seem any better now @Andysebb ?

Answer 19

What did you use product rule on? the entire left side? And the first equation I gave you was the equation already multiplied with the integration factor.

Answer 20

Oh I see, well I didn't know that, I was only inspecting the given DE, but it doesn't matter if we did the work twice, it sure shows that the steps are right.

To your question, yes the way you read the left hand side, it calls for the product rule, you differentiate it normally. That is what the d/dt Leibniz Operator tells you to do.

Answer 21

so by product rule what would be the f and the g. product rule: derivative of f * g + derivative of g + f. So how would I set that up? t(200+t)^2 * 2(200+t)/(200+t) or something along those lines?

Answer 22

well if you watch the product rule like that, then you can view f=y and g=(200+t)^2