let's solve this separable differential equation we have 1 over theta times dy D theta it's equal
to Y times sine theta over Y squared plus 1 and we also know that Y of pi is equal to 1
okay let's go ahead to move all the Y's together and move all the Thetas together let's take you
of the data first let's first multiply D theta on both sides and you see we also have another
data in the denominator on left hand side right let's go ahead multiply data both sides as well
so that they can so and let me put a data like this right here so on the left hand side we just
have 1 times dy polio is just divided by itself on the left hand side and this is equal to 4 this let
me write it down this way you see we have Y times this over that any times data let me put the Y's
together like this first Y over Y square plus 1 and we multiply by theta times sine theta and
then the D theta after that and now I just have to multiply both side by the reciprocal of this right
to the other side so let's go ahead multiply it by Y squared or Y squared plus 1 over Y Y square
plus 1 over Y so that is now cancel this and our cancel okay and you know we can integrate both
sides now right look at this Y squared plus 1 over Y to integrate that let me just split the fraction
Y squared over Y we have Y ok and then plus 1 over Y this is what I want to look at and then we
have the dy after that and this is equal to theta times sine theta D theta now integrate integrate
let's go ahead to this first integral Y is just going to be 1/2 Y square so let me put this down
Y squared over 2 and integral of 1 over y is plus absolute value Ln of absolute value of y like this
don't worry pata plus C let's do that on the right hand side ok how can we integrate theta times sine
theta integration by parts and let's do that real quick with the DI method right here ok so I put
down D and I and then plus minus plus Street rows inside enough the input on theta right here to be
differentiated and let me integrate sine theta differentiating theta one time we get one and
then do again we get zero so we stop integrating sine theta we first get negative cosine theta and
then do it again the integral cosine theta is sine theta and pass the signs that happen now right
but then we have to maintain the symbol maintain the negative because once again the integral of
cosine theta is positive sine theta but this was negative so putting it down and now the product
of the diagonals or the answer that's really nice so first we'll have positive theta times negative
cosine theta so that's negative theta because of this negative R here data and another cosine theta
and we do negative one times this so becomes past give one time step which is just y a sine
theta and this is it because we have a zero in the D column so this is it and now we are going
to put down plus C on the right hand side I'm just going to write down plus C because I don't in this
a way for me to isolate y so our answer will have to be in the implicit form y cannot be isolated I
don't think so so this is enough ok that's it for the solving part for the integration part
next we have to suffered C let's go ahead do that so let's see refer back to this we know
when X is equal to PI Y will be 1 so plug in 1/2 here we have 1 squared 1 squared over 2 and plus
Ln absolute value of y which is 1 right here and this is going to be 3 up negative theta is PI and
then we have cosine n theta is PI and we add it with and let me put parentheses around the pipe
and then right here we have sine of theta is PI and then we add a C after that the C is all the
way yet it's nice tile any other sine or cosine functions okay 1 squared is 1 divided by 2 is
1/2 Ln of 1 is 0 so on your on the left hand side we have 1/2 this is going to give us okay what's
cosine of PI cosine PI is negative 1 negative 1 times negative PI is going to give us partly pi
plus sine of Pi this right here is 0 so which is 1/2 plus C after that and of course we can
subtract PI on both sides therefore C will be 1/2 minus pi okay and we are pretty much done all we
have to do is just plugging this into the C so let me write this down for you guys right here we have
Y squared over 2 plus Ln absolute value of y and then this is equal to negative theta cosine theta
and then we add the sine theta after that and at the end the C's or that so we add 1/2 minus
PI and once they get the C it's not inside the sine and I think this is clear if you want to
make it even more clear just put parentheses around the cosine sign this right here is it
however I have one little thing I want to add right here just one remark here we have Ln of Y
with the absolute value right by in fact we can drop the absolute value because when you have a
solution to the differential equation you have to make sure that the function is continuous if
you have the absolute value in fact not confuse and second of all you know to pick the positive
version with because you have y is equal to passes one you get to plug in positive one in here Ln of
positive one is doable it's zero right so you can just replace that with the premise if you
like this is okay and this is a continuous curve and let me just read this down right here for
you guys Ln absolute value of five this means you have two things to look for this is either Ln of
regular Y or Ln of negative Y right here if Y you know it is greater than zero you use the positive
version here Y is less than zero use the negative version because that's that zero with the negative
negative negative what keep you past it can become doable in other word if earlier we found out why
of Pi it's equal to let's say this is negative five negative ten any negative numbers right here
I would replace this with a line of negative Y okay anyways this is the most legitimate
answer but then if you put down absolute for earlier I will only take a one-point test