Question

If dy/dx = sin^2( piy/4) and y = 1 when x = 0, then find the value of x when y = 3.

Answer 1

do you know how to solve the given differential equation?

Answer 2

im not too sure how to manipulate dy/dx

Answer 3

i know i have to isolate the variables and integrate and solve for y right?

Answer 4

\[\frac{dy}{dx}=\sin^2(\frac{\pi}{4} y) \\ \frac{dy}{\sin^2(\frac{\pi}{4} y)}=dx \\ \]
like do you know how to ingrate both sides of this equation?

Answer 5

well the right side is just x but im not to sure on integrating the left side

Answer 6

do you know the derivative of cot(x)?

Answer 7

-csc^2(x)?

Answer 8

\[\frac{d}{dy} \cot(u)=-\csc^2(u) \cdot \frac{du}{dy} \\ \text{ where } u=u(y)\]

Answer 9

\[\int\limits_{}^{} \csc^2(\frac{\pi}{4} y) dy \\ u=\frac{\pi}{4} y \\ du=\frac{\pi}{4} dy \\ \int\limits \csc^2(u) \frac{4}{\pi} du\]

Answer 10

do you see how to continue

Answer 11

why did you inverse the pi/4

Answer 12

because if du=pi/4 dy
then dy=4/pi du

Answer 13

i just multiplied 4/pi on both sides to isolate dy

Answer 14

okay okay hold up slow down , why are we integrating csc^2(piy/4)

Answer 15

because that was the left hand side of your equation once you separated the variables

Answer 16

okay okay i see, give me a second to get my thoughts together

Answer 17

\[\frac{1}{\sin(x)}=\csc(x)\]

Answer 18

yea but the sine was squared wouldn't that affect that

Answer 19

so you know that equation I just mentioned holds
don't you think the equation will still hold if you square both sides

Answer 20

\[(\frac{1}{\sin(x)})^2=\csc^2(x) \\ \frac{1}{\sin^2(x)}=\csc^2(x)\]

Answer 21

okay sorry i was thinking the square would be negative because it was in the denominator

Answer 22

i see what you mean though

Answer 23

okay so we are integrating \[\csc^2(\frac{ \pi y }{ 4 })\]

Answer 24

oh you were doing substitution , i see

Answer 25

okay so back to \[\int\limits\limits_{}^{} \csc^2(\frac{\pi}{4} y) dy \\ u=\frac{\pi}{4} y \\ du=\frac{\pi}{4} dy \\ \int\limits\limits \csc^2(u) \frac{4}{\pi} du \]
should be do another substitution?

Answer 26

on what? that is just -4/pi*cot(u)+C
where u=pi/4y

Answer 27

\[\frac{ 4 }{ \pi } \int\limits (-\ln(\csc(x) + \cot(x)) + C)^2\]

Answer 28

is that right?

Answer 29

minus the integral sign

Answer 30

where did you get that answer

Answer 31

remember you yourself said the d/dx cot(x) =-csc^2(x)

Answer 32

or you could say -d/dx cot(x)=csc^2(x)

Answer 33

are these wrong then?

Answer 34

\[\int\limits\limits\limits_{}^{} \csc^2(\frac{\pi}{4} y) dy \\ u=\frac{\pi}{4} y \\ du=\frac{\pi}{4} dy \\ \int\limits\limits\limits \csc^2(u) \frac{4}{\pi} du \\ \frac{-4}{\pi} \cot(u)+C \\ \frac{-4}{\pi} \cot(\frac{\pi}{4}y)+C\]

Answer 35

nope those all look great

Answer 36

arghh i feel stupid

Answer 37

anyways we also have to integrate the other side

Answer 38

\[\frac{-4}{\pi} \cot(\frac{\pi}{4} y)+C=x\]

Answer 39

-_- i was just typing that, okay so now just isolate y right

Answer 40

your second job is to find C for the point (x=0,y=1)

Answer 41

i would not solve for y
i think that would cause more work then needed

Answer 42

besides you want to find x

Answer 43

for when y=3

Answer 44

which is the third part of the problem
third and final

Answer 45

\[C = \pi\]

Answer 46

- pi

Answer 47

\[\frac{-4}{\pi} \cot(\frac{\pi}{4} y)+C=x \\ (0,1)=(x,y) \\ \frac{-4}{\pi}\cot(\frac{\pi}{4}(1))+C=0 \\ \frac{-4}{\pi} \cot(\frac{\pi}{4})+C=0 \\ \frac{-4}{\pi}(1)+C=0 \\ \frac{-4}{\pi}+C=0\]

Answer 48

:(

Answer 49

it's okay... no frowns here!

Answer 50

alright well c = 4

Answer 51

\[\frac{-4}{\pi} \cot(\frac{\pi}{4} y)+C=x \\ \frac{-4 }{\pi} \cot(\frac{\pi}{4} y)+\frac{4}{\pi}=x\]
last step find x for when y is 3

Answer 52

well no

Answer 53

c=4/pi

Answer 54

okay let me just take a breath , my frustration is getting the better of me here

Answer 55

okay so now we just plug c into the solution and plug y = 3 and find x right

Answer 56

right

Answer 57

i will let you do that
and I will come back in check in like 5 minutes or less

Answer 58

okay im getting 8/pi

Answer 59

that sounds right

Answer 60

4/pi+4/pi
2*4/pi
8/pi

Answer 61

okay great , thanks for the help
sorry for being such a dummy :P

Answer 62

i'm going to go for tonight after this problem
sleep time
but before i go do you have any last questions on this problem?

Answer 63

you aren't a dummy
probably just frustrated and trying to learn a new subject

Answer 64

no im goo thanks again !

Answer 65

the one part is hard enough
being frustrated makes it tad harder

Answer 66

anyways peace
and i gave you a medal to award you for your effort

Answer 67

:) night