Question

Seperable ODE problem

Answer 1

Ok I solved b(i) and (ii). However, it's just the end the of part (ii) I am having problems with where it says "find the general solution of x(t) to the differential equation (1)"

Answer 2

\[\int\frac{z}{z^2+2}\text dz\]

Answer 3

I have to use my solution and subsituite back into somehow I believe.

Answer 4

have you found \(z(x)\)?

Answer 5

Yes

Answer 6

I assume I have to use that back in equation (1) somehow

Answer 7

and you want \(x(t)\)?

Answer 8

Yep in equation (1)

Answer 9

I cannot separate it.

Answer 10

...multiply by \(t\text{ime}\)

Answer 11

@UnkleRhaukus What do you mean?

Answer 12

\[z(t)=x(t)/t\]
\[x(t)=tz(t)\]

Answer 13

right?

Answer 14

@UnkleRhaukus So my general solution will be something like tz=....?

Answer 15

what did you get for z(t)=

Answer 16

i mean z(x)

Answer 17

Any ideas anyone??

Answer 18

I got this

Answer 19

I got that from seperating this

Answer 20

so

\[x(t)=t\sqrt{e^{\frac{1+c}2}-2}\]

Answer 21

I didn't get the t before the square root.

Answer 22

shall we check if this is a solution to
\[tx\frac{\text dx}{\text dt}=2t^4+t^2x^2+x^2\]

Answer 23

So if sub that into the equation it should equal zero?

Answer 24

I still gotta find x(t) from the equation (1) though :/

Answer 25

Very annoying question this is, I am 90% done with the problem!

Answer 26

\[t^2\sqrt{e^{\frac{1+c}2}-2}\sqrt{e^{\frac{1+c}2}-2}=2t^4+t^4\left(e^{\frac{1+c}2}-2\right)+t^2\left(e^{\frac{1+c}2}-2\right)\]

Answer 27

How did you get that??

Answer 28

i think i missed something

Answer 29

i put x(t) into (1) to see if it was a solution

Answer 30

I see, how I don't know that helps finds x(t). I think it's impossible to separate

Answer 31

helps to find x(t)*

Answer 32

what did you get for (i)

Answer 33

I got what it asks for.

Answer 34

This continues in (ii) when you are asked to separate it.

Answer 35

\[\int\frac{z}{z^2+2}\text dz=\int t\text dt\]

Answer 36

thats why you are asked to solve that integral first. It's make it easier.

Answer 37

Yep

Answer 38

1/2ln(z^2+2)=1+c

Answer 39

=1?

Answer 40

divide by 2, take logs, subtract 2 and square root it

Answer 41

the integral of t=1

Answer 42

\[\int t\text dt=\frac{ t^2}2+c\]

Answer 43

to integrate t , add one to the index and divide by the new index

Answer 44

flutter

Answer 45

Ok @UnkleRhaukus here's my work, apparently I don't know how to integrate...

Answer 46

\[\ln z^2+2=\ln (z^2+2)\] right?

Answer 47

Yep

Answer 48

so you can't get to you 4th line from your 3rd

Answer 49

Oh god. I am so lost with this question : [

Answer 50

\[\ln(z^2+2)=t^2+c\]
\[z^2+2=e^{t^2+c}=ke^{t^2}\]

Answer 51

also, on a side note, if you ever did wind up with \(C+2\) that is a constant so you could just write \(C_1+2=C_2\) and combine the constants

Answer 52

@UnkleRhaukus

Answer 53

do we not need the \(\pm\) upon taking the sqrt ?
(I forgot, sorry it's been a while...)

Answer 54

I don't think so. To be honest I am so confused, I still have to find x= from the equation 1 somehow...

Answer 55

i was hope the sqrt was gonna cancel with the ^2

Answer 56

hoping *

Answer 57

and when you have \[\large e^{t^2+c}\]again it is customary to combine the constants\[\large e^{t^2+c}=e^{t^2}\cdot e^c\]since \(e^c\) is a constant we can rename it k and write\[\large e^{t^2+c}=e^{t^2}\cdot e^c=ke^{t^2}\]as @UnkleRhaukus did earlier

Answer 58

remember x=zt

Answer 59

Yeah

Answer 60

so does this x solfe the original DE/?

Answer 61

I am not sure :/ Would the question become seperable if I subbed in x=zt?

Answer 62

you dont need to separate the variables

Answer 63

you just have to know what x is and what dx/dt is

Answer 64

Yeah we know what x and x' is

Answer 65

x=t

\[x=t\sqrt{ke^{t^2}-2}\]
\[\frac{\text dx}{\text dt}=?\]

Answer 66

you might have to use the chain rule a bunch of times,

Answer 67

but ive got to sleep rightaboutnow good luck