Question

I have a question about methods of solving seconds order differential equations;

Answer 1

(a)\[\frac{\text d^2y}{\text dx^2}+n^2x=0\]
(b)\[\cos^2x\frac{\text d^2y}{\text dx^2}=1\]

Answer 2

My text book suggest using this technique 1 (a)\[\frac{\text d^2y}{\text dx^2}+n^2x=0\]\[\text{let }p=\frac{\text dy}{\text dx}\]\[\frac{\text dp}{\text dx}=\frac{\text d^2y}{\text dx^2}\]\[\frac{\text dp}{\text dx}+n^2x=0\]\[\frac{\text dp}{\text dx}=-n^2x\]\[p=-n^2\int x\text dx\]\[p=-n^2\frac{x^2}{2}+c_1\]\[\frac{\text dy}{\text dx}=-n^2\frac{x^2}{2}+c_1\]\[y=\frac{-n^2}{2}\int x^2+c_1\text dx\]\[y=\frac{-n^2x^3}{6}+c_1x+d_1\]\[y+\frac{n^2x^3}{6}+cx+d=0\]
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(b)\[\cos^2x\frac{\text d^2y}{\text dx^2}=1\]\[\frac{\text d^2y}{\text dx^2}=\sec^2x\]\[\text{let }p=\frac{\text dy}{\text dx}\]\[\frac{\text dp}{\text dx}=\frac{\text d^2y}{\text dx^2}\]\[\frac{\text dp}{\text dx}=\sec^2x\]\[p=\int\sec^2 x\text dx\]\[\frac{\text dy}{\text dx}=\tan(x)+c\]\[y =\int\tan x+c\cdot\text dx\]\[y=\ln \left|\sec x\right|+cx+d\]_____________________________

Answer 3

but this method seams simpler to me

(a)\[\frac{\text d^2y}{\text dx^2}+n^2x=0\]\[\frac{\text d^2y}{\text dx^2}=-n^2x\]\[\iint{\text d^2y}=\iint-n^2x\cdot{\text dx^2}\]\[\int\text dy=\int-n^2\frac{x^2}{2}+c_1\cdot\text dx\]\[y=\frac{-n^2x^3}{6}+c_1x+d_1\]\[y+\frac{n^2x^3}{6}+cx+d=0\]

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(b)\[\cos^2x\frac{\text d^2y}{\text dx^2}=1\]\[\frac{\text d^2y}{\text dx^2}=\sec^2x\]\[\iint\text d^2y=\iint\sec^2x\cdot\text dx^2\]%\cdot\text dx\]\[\int \text dy =\int\tan x+c\cdot\text dx\]\[y=\ln \left|\sec x\right|+cx+d\]______________________________

Answer 4

{ignore that line starting with %}

Answer 5

is the second method valid. ?

Answer 6

n is some constand?

Answer 7

and why not laplace transforms on the second?

Answer 8

yes n is just a constant term

Answer 9

yes the second form is valid

Answer 10

where does d come from, im confused

Answer 11

o i c

Answer 12

d is the second constant of integration

Answer 13

or do you mean the infintesimal change \(\text d x\)

Answer 14

nope, i see now.

Answer 15

I would just use laplace but this all seems cool:)

Answer 16

i think it will be valid for all equations in form:
\[d^2y/dx^2+f(x)=0\]

Answer 17

your way on a is quite neet, good job

Answer 18

i cant seam to find any reference to solving the DE's using the second method,

Answer 19

yeah no idea, seems like if it would work we would have seen it. lil scetchy but the "proof" is in the pudding

Answer 20

how about:
\[d^2y/dx^2+f(x)g(y)=0\]

Answer 21

thats seperable in the same since so it would prob work?

Answer 22

like:
\[d^2y/dx^2+x^2/y^2=0\]

Answer 23

\[\frac{\text d^2y}{\text dx^2}+f(x)g(y)=0\]\[\frac{\text d^2y}{\text dx^2}=-f(x)g(y)\]\[\frac{\text d^2y}{g(y)}=-f(x){\text dx^2}\]

\[\iint\frac{\text d^2y}{g(y)}=\iint-f(x){\text dx^2}\]

Answer 24

yep

Answer 25

damn your fast with that equation editor, i was on line two

Answer 26

i actually am typing by hand

Answer 27

I will show this to my dif eq prob on monday:) or just use it in a test, ha

Answer 28

how?

Answer 29

o I dont know that code ") and im lazy

Answer 30

prob*prof ?

Answer 31

prof*

Answer 32

I dont know how to do the editor well but int(int(dy),dy) would this not be y^2/2+cy+d?

Answer 33

do i have to use convolution to take the Laplace transform of the second equation?

Answer 34

yeah

Answer 35

actually
you could just find the laplace of sec^x

Answer 36

err sec^2(x)

Answer 37

nm that does not look fun

Answer 38

i cant remember how to find the laplace transform of that

Answer 39

by deffinition int(sec(x)^2*e^(-s*t),t,0,infinity)

Answer 40

but thats looks very lame

Answer 41

concolution is more about the laplace inverse of a product of laplaces i think

Answer 42

convalution*

Answer 43

ha

Answer 44

but what about that double integral thing I mentioned?

Answer 45

um that is what i first though
http://openstudy.com/users/unklerhaukus#/updates/4fe4a333e4b06e92b8730228

Answer 46

yeah seems a little odd. but hey why not

Answer 47

seems a little to nice a coincidence tou got the right answer both times

Answer 48

tou = you

Answer 49

ahh

Answer 50

notice how these are different \[\iint\text d^2y = \iint \text d(\text d y)=\int \text dy=y+c\]this is different to\[\iint\text dx^2 = \iint \text dx\text d x=\int x+c \text dx=\frac{x^2}{2}+cx+d\]

Answer 51

yeah I see whawt your saying

Answer 52

is there any example you can think of where my second method fails

Answer 53

nope and I tried like 5 of my questions dealing with integrating factors and seperable equations. But it seems like something must be wrong lol. either way nice work

Answer 54

How would you integrate
\[\int\limits\int\limits \frac{d^{2}y}{g(y)} \]
in the example that \[\frac{d^{2}y}{dx^{2}} = f(x)g(y)\]

Answer 55

\[\iint\frac{\text d^2y}{g(y)}=\iint-f(x){\text dx^2}\]for \(g(y)=1/y^{2}\) \(f(x)=x^2\)
\[\iint y^2\text d^2 y =-\iint x^2{\text dx^2}\]

Answer 56

Your method does seem to work, Rhaukus. As far as I can tell, the substitution is absolutely valid by definition.

I cannot say with 100% confidence it works all the time, but I am 95% confident it does.

Answer 57

so what is \[\int\limits\int\limits y^{2}d^{2}y\]
I really dont see how I can evaluate that given the fact that this doesnt make any sense to me

Answer 58

read that link he posted to me

Answer 59

its seems undefined method but seems to work?

Answer 60

I read that discussion yesterday. I am still in the dark as to how I would integrate y^2d(d(y)).... doesnt make any sense to me at the moment

Answer 61

\[\iint y^2\text d \text d y =-\int \left(\int x^2{\text dx}\right)\text dx\]
let\[u=\text {d}y \]\[\frac{u^2}{2}=y\]\[\text du=\text {dd}y\]

\[\int u^2\text du=-\int \frac{x^3}2+c\cdot \text dx\]\[ \frac{u^3}2 =-\frac{x^4}{4} +cx+d\cdot \text dx\]

?

Answer 62

should be \(u^3/3\) on the LSH

Answer 63

i mean \(\frac{u^3}{12}\)

Answer 64

does that make sense to anyone/

Answer 65

i dont think the equations i just wrote make any sense anmore

Answer 66

i think ill just stick to solving these by treating them as y absent second order DEs

Answer 67

hey....both the methods are same as u r integrating two times in these methods.....and so u get the same ans...

Answer 68

\[\iint \text d^2y\]still dosent feel right

Answer 69

wat does it mean?????

Answer 70

i just can't visualize \(\text d^2y=\text {dd}y\)
the infi-infinitesimal .

Answer 71

∬d2y is always equals to y ,because u are finding the double integral of double derivative...

and as integration is the inverse of differentiation.....we get the ans.. as y...

Answer 72

so is this right then? \[\iint y^2\text d \text d y =\frac{y^4}{12}\]

Answer 73

I am not convinced in this step:
u = dy
u^2/2 = y
Aren't we supposed to like integrate wrt to something ? for LHS
I mean in
\[\int\limits u = \int\limits dy\]
how can can you write \[\int\limits u = u^{2}/2\] considering that you arent integrating u wrt anything.

Answer 74

sorry for the late reply ... had gone for lunch

Answer 75

yes...