Question

vdv/dx = 2(16+v^2) differential equation

Answer 1

it is a fully seperatable DE so you should put all the v terms on one side and all the x terms on the other

Answer 2

that's what i cant do, because there's v on both sides...

Answer 3

do i divide by 2(16+v^2)?

Answer 4

yes.

\[\frac{ v dv }{ (16 + v^{2} ) } = 2 dx\]

Answer 5

but how do i integrate the left hand side?

Answer 6

is it 2vln(16+v^2)

Answer 7

let u = 16 + v^2 --> du = 2v dv

= \[1/2 \int\limits_{}^{} \frac{ du }{ u }\]

Answer 8

= 1/2 ln | 16 + v^2 | + C

Answer 9

hmmm did u do that by substitution???? i dont normally do it that way...

Answer 10

technically you don't have to since it can be recognized and there are definitely other valid approaches. but i find it to be the simplest way to look at it.

Answer 11

it was 1/2 ln |u|. plugged back in u = v^2 + 16

Answer 12

how did we get this bit? \[1/2 \int\limits_{?}^{?} u/du\]

Answer 13

sorry :( my integrations really bad x

Answer 14

we had \[\int\limits_{}^{} \frac{ v dv}{ (v^{2} + 16) }\]

u = v^2 + 16 ----> du = 2v dv <-----> v dv = (1/2) du

= \[\int\limits_{}^{} \frac{ (1/2) du }{ u }\]

Answer 15

aaaaaaaaahhhhhhhhhhhhh i seeeeee. give me one moment to assiimilate that thought plz

Answer 16

what about the v on the top of the integral??? where did that go?

Answer 17

v dv = 1/2 du

the substitution was chosen for that to happen

Answer 18

ahh ok :) and the right hand side easily integrates to 2x ???

Answer 19

so

Answer 20

1/2 ln (16 +v^2) = 2x +c

Answer 21

yes

Answer 22

then

Answer 23

ln(16+v^2) = 4x +c

Answer 24

yes

Answer 25

take e of both sides?

Answer 26

yes

Answer 27

oh wait i forgot to mention that x=0 when v=0, so I need to find C first

Answer 28

you don't have to find it yet. usually easier to find once you have you v(x) function at the end

Answer 29

oh okay i will try that now then :)

Answer 30

this next part is actually a little tricky

Answer 31

oh. 16+v^2 = e^4x +c

Answer 32

??????

Answer 33

taking e of both sides gives:

16 + v^2 = \[e^{4x + C } = e^{4x} e^{C}\] by law of exponents

Answer 34

e^C will always give a real poisitive number so we can say that e^C = C. giving Ce^(4x)

Answer 35

yeah i get the first bit about law of exponents, but i dont get the second bit ...

Answer 36

oh wait...

Answer 37

the same way that when you multiplied both sides by 2 you got 4x + C instead of 4x + 2C. 2C is pretty redundant since C is an arbitrary constant that we don't really care how it was obtained. e^C can give any positive number in the world. same way that C can be any number in the world

Answer 38

oh yeaaaahhh.... i see :)

Answer 39

so

Answer 40

erm, now what do we do., we have now got

Answer 41

16+V^2 = Ce^(4x

Answer 42

not quite sure how to find C

Answer 43

(i am now your fan btw :))

Answer 44

C will be the last thing you find suing the initial condition v(0) = 0.

you want to isolate v. this will give you a function v(x) which is the solution to the differential equation

Answer 45

cool :)

Answer 46

using*

Answer 47

erm....speak english??? sorry no comprendo last sentence

Answer 48

lol. you need to solve for v. isolate v bu itself on one side

Answer 49

ah

Answer 50

i take 16 over to the other side of eqn, then sqrt to get v=.....

Answer 51

yes

Answer 52

\[v=\sqrt{Ce ^{4x} -16}\]

Answer 53

yes.

now you know that at x = 0. v = 0
[ v(0) = 0 ]

Answer 54

this will give you C which is unique to that initial condition

Answer 55

0=sqrt(C-16)

Answer 56

ya

Answer 57

C^(1/2) = 4

Answer 58

c=16?

Answer 59

ya c = 16

Answer 60

so

Answer 61

ln(16+V^2) = 4x + 16 ?????

Answer 62

oh actually no

Answer 63

v=?

Answer 64

the final equation is the solution to your differential equation

yes v(x)

Answer 65

\[v=\sqrt{16e ^{4x }-16}\]

Answer 66

yes. it can be simplified a bit nice but it is correct

Answer 67

v= 4e^(4x) -4

Answer 68

urghhhh... why did you take the 4 outside the sqrt?

Answer 69

oh yeah yeah yeah right sorry yeah

Answer 70

so

Answer 71

i think

Answer 72

v= 4 (sqrt e^(4x)) -4

Answer 73

oh this is getting tricky

Answer 74

erhhh... that's the answer i think??? :)

Answer 75

\[v(x) = 4\sqrt{e^{4x} - 1}\]

Answer 76

what i said earlier may have been misleading. deleted it

Answer 77

\[\sqrt{a}*\sqrt{b} = \sqrt{ab}\]

\[\sqrt{16e^{4x} - 16} = \sqrt{16}\sqrt{e^{4x} - 1}\]

Answer 78

ah Euler, you are a genius!!!! wow

Answer 79

i finally got it.

Answer 80

wow, thank u soo much. i am revising for A levels. its very tricky, but you are so very helpful :)

Answer 81

i love this website too

Answer 82

i'm glad i could help :) i love math and helping

Answer 83

you're doing a great job helping loads of people all over the world. superman. I really appreciate it

Answer 84

Thank you for your patience and everything :) <3 i will definitely come back again soon

Answer 85

glad to hear that :) enjoy math

Answer 86

r u a university graduate btw???

Answer 87

or a professor of math or something?

Answer 88

currently an undergraduate.[bachelor's then leaving]

Answer 89

no. i just love math. i tutor friends

Answer 90

oh wow. oh ah ok. its really good that u love math. keep spreading the love.

Answer 91

hehe will do :)

Answer 92

ok shall i go now...?

Answer 93

yeah. i think ill give it a rest now. Thank you sooo much.

Answer 94

see you . many thanks. bye xxxxxxxxx