Question

Second Order ODE
\[\cos^2x\frac{\text d^2y}{\text dx^2}=1\]

Answer 1

\[\cos^2x\frac{\text d^2y}{\text dx^2}=1\]\[\frac{\text d^2y}{\text dx^2}=\sec^2x\]\[\iint\text d^2y=\iint\sec^2x\cdot\text dx^2\]\[\int y\cdot\text dy =\int\tan x+c\cdot\text dx\]\[\frac 12y^2=\ln \left|\sec x\right|+cx+d\]

Answer 2

my book has a different answer to this have i missed something?

Answer 3

what the heck does dx^2 mean o-0 ?

Answer 4

\[\cos^2x\frac{\text d y \text dy}{\text dx\text dx}=1\]\[\frac{\text d y \text dy}{\text dx\text dx}=\sec^2x\]\[\iint\text dy\cdot\text dy=\iint\sec^2x\cdot\text dx\cdot\text dx\]\[\int y\cdot\text dy =\int\tan x+c\cdot\text dx\]\[\frac 12y^2=\ln \left|\sec x\right|+cx+d\]

Answer 5

and your answer is right on the RHS, but that 1/2y^2 is wrong

Answer 6

that is an ill-formed way to write it in my opinion
I would just take it one step at a time

Answer 7

try differentiating it twice ... if you get the DE then it's correct.

Answer 8

how come the LHS is just \(y\)?

Answer 9

if there is a simpler way what is it/

Answer 10

I would write
\[\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{dy}{dx})\]
Then integrate both sides wrt x

Answer 11

\[\cos^2 xy''=1\]\[\frac{d^2y}{dx^2}=\sec^2x\implies\frac{dy}{dx}=\int\sec^2x dx=\ln|\sec x|+C_1\]seperation of the variables on the second derivative will not give you y, but y' instead

Answer 12

@FoolAroundMath has stated the formality better that I...

Answer 13

oops, got ahead of myself above :/\[\cos^2 xy''=1\]\[\frac{d^2y}{dx^2}=\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{dy}{dx})=\sec^2x\implies\frac{dy}{dx}=\int\sec^2 x dx=\tan x+C_1\]separate the variables again\[y=\int\tan x+C_1dx=\ln|\sec x|+C_1x+C_2\]

Answer 14

@FoolAroundMath 's notation should make it more clear why we don't get 1/2y^2 on the LHS

Answer 15

wolf says correct
http://www.wolframalpha.com/input/?i=y%27%27+-+sec^2x+%3D+0

Answer 16

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

Answer 17

right, and as I mentioned I think @FoolAroundMath 's notation makes the process even more clear

Answer 18

a \(\text d\) magically melts away

Answer 19

no it doesn't, watch closely...

Answer 20

\[\cos^2x\frac{\text d^2y}{\text dx^2}=1\]\[\frac{\text d^2y}{\text dx^2}=\frac d{dx}\left(\frac{dy}{dx}\right)=\sec^2x\]\[\int d\left(\frac{dy}{dx}\right)=\int\sec^2x dx\]\[\frac{dy}{dx}=\tan x+C_1\]\[\int dy=\int \tan x+C_1dx\]\[y=\ln|\sec x|+C_1x+C_2\]

Answer 21

notice that by utilizing the notation in the second line it becomes clear that separation of variables is being applied as usual

Answer 22

I have to wait for this to sink in \[\int \text d\left(\frac{\text dy}{\text dx}\right)=\frac{\text dy }{\text dx}\]

Answer 23

let\[u=\frac{dy}{dx}\]is not\[\int du=u+C\]?

Answer 24

we could write this whole thing subbing u for dy/dx

Answer 25

it make perfect sense with a subsitution
\[\int \text du=u+c\]

Answer 26

yeah it just took a while to get it

Answer 27

I'm just learning a lot watching this debate ^_^ Perhaps a tutorial on the proper syntax usage for all these \(\frac{d^2x}{dx^2}\) & \(\frac{d}{dx}\) & dx types of things would be in order? (gives @FoolAroundMath a medal for helping out too)

Answer 28

does\[\iint \text d^2y=y\]

Answer 29

+c

Answer 30

yeah, I learned from the user JamesJ (an actual professor), that \[\int\int d^2y\]actually has no meaning
it's undefined, and is what is called an ill-formed expression

Answer 31

oh,

Answer 32

along with\[\int\int(dx)^2;~~~\int\int dxdx\]and other things like that
in short, you do have to be very careful with the notation in cases like this

Answer 33

why is that ?

Answer 34

well, what would it mean? nobody has given it one

Answer 35

if you want to invent a meaning for those expressions above, that's fine, but it won't give you the right answer to your ODE in the way that you have chosen to define it

Answer 36

it would have worked if
\[\iint \text d^2y=y\]

Answer 37

but you have to analyze what \(d^2y\) means
is it \(d(dy)\) ?

Answer 38

yes

Answer 39

oh it could be dydy

Answer 40

i see the confusion

Answer 41

this is pushing the boundary of my knowledge, but I know for sure that dydy is an ill-formed expression
as far as d^2y=d(dy) I'm not sure about that, but if so I guess that would mean that\[\int\int d^2y=\int\int d(dy)=\int dy=y+C\]but then what do you do with \[dx^2\]?
is that\[dx^2=d(dx)\]I don't think that works...

Answer 42

\[\int f(x)dx^2;~~~~~\int\int f(x)dx^2\]don't mean anything to me

Answer 43

\[\text d^2w=\text d\cdot\text dw\]
\[\text dw^2=\text dw\cdot\text dw\]

Answer 44

$$dx^2 \text{ looks more like } d(x^2) = du \text{ under substitution } u =x^2$$

Answer 45

@UnkleRhaukus those definitions I suppose make your ODE come out right, so go ahead and use them I guess. I don't know if those are generally accepted by the mathematical community
either way, you see how that by defining \(d^2y=d(dy)\) <_(I really think this one should have parentheses to show that d(dy) is the differential of the differential, and no some multiplication)
and as I said, I'mm pretty sure that there is some mathematical subtly sin committed by writing\[\int\int f(x)dxdx\]but I am still a student, and would need clarification as to why that is the case

Answer 46

http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio
i guess this would help

Answer 47

better perhaps if we could find an entry on the possibility of using dxdx for dA in a double integral