Question

Integration help..

how does (d^4)y/dx^4 = w / EI gives ( when integrated 4 times)

y(x)=c1 + c2x +c3x^2 + c4x^4 + (w/IE)x^4

I just can't see it... any help>?

Answer 1

can you define w, E, l ?

Answer 2

are they constants?

Answer 3

I can't its for beam deflection and this is just a general equation, but the question is based on it. w is a constant, so is e and I , just not defined in my case

Answer 4

kk, constant is all I need to know

Answer 5

so, if you have \[\int x\] what does that equal?

Answer 6

xdx sorry

Answer 7

@MarcLeclair

Answer 8

its x^2/2

Answer 9

+ c for undefined integrals

Answer 10

ah, but you are missi-

Answer 11

yes, plus c. good

Answer 12

now how about \[\int 1 dx\]

Answer 13

mhmmm x + c

Answer 14

sorry taking long to answer, doing 2 things at the sametime

Answer 15

good, now \[\int (x+c)dx\]

Answer 16

stay with me for 5 minutes and you'll be done

Answer 17

x+c will give you x^2/2 + c and x + c

Answer 18

well not quite

Answer 19

so x^2/2 yes, but just x, no

Answer 20

you need to integrate \[\int C~dx\]

Answer 21

what happens to a constant when we integrate?

Answer 22

c would be a constant, like 1 , so x + c no?

Answer 23

no, think, if you have \[\int 3x ~dx\] What do you get?

Answer 24

i would get 3x^2/2

Answer 25

good, so what did you do with the 3?

Answer 26

just factored it out

Answer 27

then multiplied it back right?

Answer 28

yeah haha

Answer 29

so you can't just drop your constants when you integrate something like C

Answer 30

so, what is the integral of C dx now?

Answer 31

it would be cx

Answer 32

cx + what?

Answer 33

cx + c^2 righjt?

Answer 34

c ( x + c)

Answer 35

no, it's Cx +(some new constant)

Answer 36

we add a different constant each time

Answer 37

wait, because we have c(int x dx ) wouldnt it be c ( x^2/2 + c)

Answer 38

nah, we did \[C\int 1dx\]

Answer 39

and remember we do a new constant every time, also you have to distribute to every value inside the parentheses

Answer 40

That's why we end up with \(C_1,C_2,C_3\)

Answer 41

yeah so I first integrate 1, it should give me x + c , however in this case its always multiplied ( ie x*c2 ) which is confusing to me

Answer 42

noooo you do not get x+c

Answer 43

you get \(C_1x+C_2\) or \(Cx+a\) whatever you wanna add, but you have to keep your constant

Answer 44

mhm so if I start with int ( d^4y/ dx^4) i would then get c1 d^3y/d^3x then its going to be c1x + c2? and so on?

Answer 45

just like when you had 3, you still got 3/2x^2 +c the three didn't go away

Answer 46

uhh, you don't go about it that way

Answer 47

but I thik you have the right idea, but your first constant is your w/el thingy

Answer 48

So its just a constant call it Z yea? you do \[\int Zdx=ZX+C_1\]

Answer 49

\[𝑦(𝑥) = 𝑐1 + 𝑐2𝑥 + 𝑐3𝑥 2 + 𝑐4𝑥 3 + 𝑤0 24𝐸𝐼 𝑥 4\] that would be the complete answer.

Answer 50

woahhh sorry formating. \[𝑦(𝑥) = 𝑐1 + 𝑐2𝑥 + 𝑐3𝑥 2 + 𝑐4𝑥 3 + 𝑤 /24𝐸𝐼 (𝑥 4)\]

Answer 51

close

Answer 52

h/o

Answer 53

yes, that is correct

Answer 54

(I was looking at the original) Now they just decided I don't want there to be a 24 near my big fraction constant so I undistribute the 24 and have \((C_4+Z)\)

Answer 55

\((C_4+Z)x^4\) ***

Answer 56

mhm,. the thing that throws me off is the fact that we started with the constant w/IE and we end up with x^4 in the same term. To me it goes something like:

int ( d^4y/dx^4) = w/IE , I would use DE to solve that, no?

Answer 57

no, it's an integral on both sides

Answer 58

it's \[\frac{d^3y}{dx^3}=\int Zdx\]

Answer 59

then \[\frac{d^2y}{dx^2}=\int \int Zdx\]

Answer 60

should be another dx there, but you get the gist

Answer 61

I see... then that would give you something like Z ( x + c) for the first integral? and so on?

Answer 62

I do get the gist though :)

Answer 63

distribute the Z

Answer 64

That is very important to do

Answer 65

in order to get that particular general form that is

Answer 66

so then Im left with zx + zc where zc can be C3 so then id have int zx + c3. this would give me then z(x+ c) + c3(x+c) and so on?

Answer 67

x^2/2*

Answer 68

you are adding too many constants

Answer 69

you need to add one per integration or combine them when possible

Answer 70

and no, you can't get rid of z

Answer 71

so first integral yeilds \[zx+c_1\] second we get \[zx^2/2+c_1x+c_2\]

Answer 72

now you do 3rd and fourth

Answer 73

thirs would be zx^3/6 + c1x^2/2+c2x + c3

Answer 74

good

Answer 75

and fourth?

Answer 76

zx^4/24 + c1x^3/6+ c2x^2/2+ c3x + c4.

Answer 77

Thanks for your patience man, super appreciated.

Answer 78

now, absorb those extra numbers into the respective constants and separate the x^4 constants and you are done

Answer 79

np, happy you got it