Question

Stuck on very last part of these 2 questions!

Answer 1

any thoughts ?

Answer 2

\[y=\cos(m \arcsin(x))\]
\[\text{ Let } u=m \arcsin(x) \\ \frac{u}{m}=\arcsin(x) \\ \sin(\frac{u}{m})=x \\ \cos(\frac{u}{m}) \frac{du}{m}=dx \\ \frac{1}{m}\cos(\frac{u}{m})=\frac{dx}{du} \\ u_1=\frac{du}{dx}=m \sec(\frac{u}{m}) \\ \text{ so anyways } \\ y=\cos(u) \\ y_1=-\sin(u) u_1 \\ y_2=-\sin(u) u_2 -\cos(u) u_1 u_1=-\sin(u) u_2-\cos(u)u_1^2\]
\[y_3=-\cos(u)u_2u_1-\sin(u)u_3+\sin(u)u_1u_1^2-2\cos(u)u_1u_2 \\ y_3=-\cos(u)u_2u_1-2\cos(u)u_1u_2-\sin(u)u_3+\sin(u)u_1^3 \\ y_3=-3\cos(u)u_2u_1-\sin(u)[u_3-u^3_1]\]
\[y_4=-3[-\sin(u)u_1u_2u_1+\cos(u)u_3u_1+\cos(u)u_2u_2] \\- \cos(u)u_1[u_3-u_1^3]-\sin(u)[u_4-3u_1^2u_2]\]
\[y_4=-3\cos(u)u_3u_1-3\cos(u)u_2^2-\cos(u)u_1u_3+\cos(u)u_1^4\\ -\sin(u)u_4+3\sin(u)u_1^2u_2+3\sin(u)u_1^2u_2 \]
\[y_4=-4\cos(u)u_3u_1-3\cos(u)u_2^2+\cos(u)u_1^4-\sin(u)u_4+4\sin(u)u_1^2u_2 \]

\[y_4=\cos(u)[-4u_3u_1-3u_2^2+u^4_1]-\sin(u)[u_4-4u^2_1u_2] \]

....I was trying to find y_n this way
but I do not see a pattern

Answer 3

once i found y_n i was going to but it back in terms of x instead of u

Answer 4

i see mention of lebnitz theorem
i either do not recall that theorem or remember

Answer 5

i need to use it...
(saw similar other questions using it)
let me pull out that theorem ...

Answer 6

at least i can still recall my cal 1 skills lol

Answer 7

http://wdjoyner.com/teach/calc1-sage/html/node106.html

Answer 8

yeah, i thought i made some mistake in finding 2nd derivative and got one sqrt (1-x^2) extra...but could not find the error....if there's no error in y_2 then, lebnitz also is not giving the required result

Answer 9

i guess induction would not work
i haven't tried it

Answer 10

and i don't know at this point if it sounds dumb or not to you
but i might look at it

Answer 11

\( (1-x^2 ) y_2-xy_1+\dfrac{(m^2 y)}{√((1-x^2 ) )}=0\)

got this
if only i would have got
\( (1-x^2 ) y_2-xy_1+ m^2 y =0 \)
then i would have got the required result

Answer 12

is my quotient rule correct ?

Answer 13

i guess i just missed the sqrt (1-x^2) in the numerator...!?

Answer 14

i have

Answer 15

but i should have

Answer 16

\[(1-x^2) y^2 \\ =-(1-x^2) \cdot \frac{[xmsin(m \arcsin(x))+\sqrt{1-x^2}m^2\cos(m \arcsin(x))]}{1-x^2} \\ =-[xmsin(m \arcsin(x))+\sqrt{1-x^2}m^2\cos(m \arcsin(x))] \]

Answer 17

yeah! thats it...i didn't apply quotient rule properly! :P
and wasted both of our time :O
*shame on me*

please go through next question now....

Answer 18

\[(1-x^2) y_2 \\ =-(1-x^2) \cdot \frac{[xmsin(m \arcsin(x))+\sqrt{1-x^2}m^2\cos(m \arcsin(x))]}{1-x^2} \\ =-[xmsin(m \arcsin(x))+\sqrt{1-x^2}m^2\cos(m \arcsin(x))]
\] \[
-xy_1 \\ =x \cdot \frac{m \sin(m \arcsin(x))}{\sqrt{1-x^2}}\]

Answer 19

stupid question for next question
what is the objective

Answer 20

i proved, using that theorem, that the value of that big expression is
\(n^2 u\)

but that is for homogeneous equation of the order n

i have \(u=e^{(x+y)}+\log (x^3+y^3-x^2 y-xy^2)\)

how will i prove that it is homogeneous with order 3
if its is that

Answer 21

it might not be a big deal to you but i see something i don't like with your rewriting of u in your word document
you probably already caught it
but it was the last line in the text you put here
\[e^{x+y}=e^{\frac{1}{x}(1+\frac{y}{x})} \neq e^\frac{1}{x} \cdot e^{1+\frac{y}{x}}=e^{\frac{1}{x}+1+\frac{y}{x}}=e^\frac{1+x+y}{x}\]

Answer 22

oops and i messed up to lol

Answer 23

oops, i thought
\(a^{mn} = a^m a^n \)

Answer 24

\[e^{x+y}=e^{x(1+\frac{y}{x})}\]

Answer 25

\[a^m \cdot a^n=a^{m+n} \text{ and } (a^m)^n=(a^n)^m=a^{nm}\]

Answer 26

right
got that mistake
but how to show that its homogeneous, if it is...

Answer 27

i was able to follow everything above that
but until we got u=e^whatever+log(stuff)
was this u given to you or did you find this u from your solving

Answer 28

given in the question

Answer 29

ok great
i was like @hartnn is so smart

Answer 30

I wonder if @tkhunny would know
I need to do a little more reading I think

Answer 31

so my main doubt is
\(u=e^{(x+y)}+\log (x^3+y^3-x^2 y-xy^2)\)

how would i prove that it is homogeneous equation of order 3
if it is that

Answer 32

"A partial differential equation is an equation that involves an unknown function and its partial derivatives. The order of the highest derivative defines the order of the equation. The equation is called linear if the unknown function only appears in a linear form. Finally, the equation is homogeneous if every term involves the unknown function or its partial derivatives and inhomogeneous if it does not. "

Answer 33

I copied this
it says every term involves the unknown function or its partial derivatives

Answer 34

Leibnitz requires FINITE derivatives?

Answer 35

that question is solved....
please look through the next one!

Answer 36

unknown functions here are
log and exponential ?

and since they are linear
u is homogeneous ?

Answer 37

I copied that from this link
http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node10.html
I

Answer 38

that says
\(\ln (x^2+y^2)\)
is homogeneous, right ?
but with what order ?
2 ?

Answer 39

i have the theorem,
i have the function
just don't know how to apply it

as i need to prove that u is homogeneous first

and want to confirm that the order is 3
so that final answer becomes 9u...

Answer 40

let me ask that as a separate question....

Answer 41

Have you considered factoring the argument of the logarithm, creating a much simpler expression whose partial derivatives fail to be particularly creative?

Answer 42

2 ln (x-y) + ln (x+y)

Answer 43

how does that help ?

Answer 44

If it's 3rd order, with two independent variables x, and y, we shouldn't get TOO many variations of partial derivatives. Those simpler expressions seemed at least a little promising.