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

- anonymous

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

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- freckles

do you know how to solve the given differential equation?

- anonymous

im not too sure how to manipulate dy/dx

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- freckles

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

- anonymous

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

- freckles

do you know the derivative of cot(x)?

- anonymous

-csc^2(x)?

- freckles

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

- freckles

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

- freckles

do you see how to continue

- anonymous

why did you inverse the pi/4

- freckles

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

- freckles

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

- anonymous

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

- freckles

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

- anonymous

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

- freckles

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

- anonymous

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

- freckles

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

- freckles

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

- anonymous

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

- anonymous

i see what you mean though

- anonymous

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

- anonymous

oh you were doing substitution , i see

- anonymous

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?

- freckles

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

- anonymous

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

- anonymous

is that right?

- anonymous

minus the integral sign

- freckles

where did you get that answer

- freckles

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

- freckles

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

- anonymous

##### 1 Attachment

- anonymous

are these wrong then?

- freckles

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

- freckles

nope those all look great

- anonymous

arghh i feel stupid

- freckles

anyways we also have to integrate the other side

- freckles

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

- anonymous

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

- freckles

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

- freckles

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

- freckles

besides you want to find x

- freckles

for when y=3

- freckles

which is the third part of the problem
third and final

- anonymous

\[C = \pi\]

- anonymous

- pi

- freckles

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

- anonymous

:(

- freckles

it's okay... no frowns here!

- anonymous

alright well c = 4

- freckles

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

- freckles

well no

- freckles

c=4/pi

- anonymous

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

- anonymous

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

- freckles

right

- freckles

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

- anonymous

okay im getting 8/pi

- freckles

that sounds right

- freckles

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

- anonymous

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

- freckles

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?

- freckles

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

- anonymous

no im goo thanks again !

- freckles

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

- freckles

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

- anonymous

:) night

Looking for something else?

Not the answer you are looking for? Search for more explanations.