Question

First order DE question

http://i.imgur.com/HHW2M.jpg

Answer 1

Do I just sub x/t for z?

Answer 2

lana you go for this easy one :P

Answer 3

\[\huge{x=zt}\]now substitute that in the first equation

Answer 4

\[t*(zt)dx/dt=2(zt)^{4}+t ^{2}(zt)^{2}+(zt)^{2}\]

Answer 5

now divide by \[\huge{t^2}\]

Answer 6

wait 2(zt)^4 is not right theres no x there

Answer 7

oh crap sorry, lemme fix that

Answer 8

t∗(zt)dx/dt=2t^4+t^2(zt)^2+(zt)^2

Answer 9

(1/t)(z)(dx/dt)=\[2t ^{2}+z^{2}+z ^{2}\]

Answer 10

I assume I multiply both side by t?

Answer 11

x=zt
x'=z't+z
\[\frac{dx}{dt}=\frac{dz}{dt}+z\]
so substtituting this in the equation
\[t^2 z (\frac{dz}{dt}+z)=2(t)^4+t^2(zt)^2+(zt)^2\]\[z(\frac{dz}{dt}+z)=2t^2+z^2t^2+z^2\]\[z \frac{dz}{dt}+z^2=2t^2+z^2t^2+z^2\]subtract z^2 from both sides\[z \frac{dz}{dt}=2t^2+z^2t^2\]\[z \frac{dz}{dt}=t^2(z^2+2)\]

i think i went wrong somewhere when simplifying i have an extra t lol

Answer 12

i cant find any mistakes in my solution i dont know

Answer 13

How did you get x'=z't+z

Answer 14

Looks the product rule but, what are your values you used to for it?

Answer 15

u want to make the whole equation in terms of Z and t
so
x=zt
now taking the derivative of bot sides with respect to t
\[\frac{dx}{dt}=\frac{dz}{dt}t+z(1)\]

Answer 16

now i know where i went wrong

Answer 17

one min let me fix it

Answer 18

The product rule won't work for out me...

Answer 19

On a side note, this is solvable using the Bernoulli method because it's a Bernoulli equation :-)

Answer 20

I can't believe I am getting confused about the product rule.

Answer 21

I have no idea how lalay got that.

Answer 22

derive both by sides in respect to t I understand but where does the +z come from.

Answer 23

Because x = zt, right? Remember the product rule: \[ \frac{d(vu)}{dx} = v \frac{du}{dx} + u \frac{dv}{dx} \]So, d(zt)/dx = \[ z \frac{dt}{dt} + t \frac{dz}{dt}\]What is dt/dt?

Answer 24

Typo, should be d(zt)/dt

Answer 25

1!

Answer 26

So, we are left with z + t z', as wished.

Answer 27

Just to confirm you're letting u=t v=z

Answer 28

Yeah, either all should work fine, as we take the derivative with respect to t, the dt/dt part will always be 1. :-)

Answer 29

Thank you for all the help, I suck at this!

Answer 30

@ironictoaster
I seriously can't believe that you asked this exact same question again.
http://openstudy.com/study#/updates/4fa6c8e8e4b029e9dc36452e

Answer 31

@estudier

I only asked again because that question was apart of a project I needed to upload as apart as a group. There's 4 in a group and we all did the question individually and we all had slightly different answers, I needed a second opinion cause the project it worth 70% of my final grade.