let's solve this separable differential equation we have one-half dy/dx is equal to square root of y plus 1
times cos x and we also know why of pi is equal to 0 so be sure to suffer that constant and I won't
have to talk about when I wish yourself with a constant in this video ok let's go ahead and do
our usual business let's go ahead multiply the x on both sides and I've also had to divide both
side by square root of y plus 1 and left-hand side let me just put this times 1 over square root of
y plus 1 like this okay to clean 6 up on the left hand side 1 times 1 is just 1 well ready times 1
over for the twin front 1 over 2 square root of y plus 1 and we have a deal why here and that's
equal to on the right hand side which is of course X X and we're ready to integrate integrate right
because all the Y's and all the eggs behind the right place what's the integral 1 over 2 squared
of y plus 1 this right here is just going to be square root of y plus 1 because a few ways you
can do it of course you can use u substitution that U is equal to y plus 1 go to initial step
right but I will suggest you guys to do it this way do kidding is that and you know that the root
of y plus 1 is just 1 so I don't have to divide or multiply by any constants and you know the general
form 1 over 2 square of something it's going to keep you score something okay and the best way
to do it is you integrate this in your head put on an answer just like this and differentiate the
answer and see if you can get back to this okay let's go ahead do that differentiate square root
of y plus 1 we first get 1 over 2 square root of the inside right y plus 1 and you multiply
by the 2 repeat is that by the chain rule the review of y plus 1 just 1 and you get back to
the original okay so right here if this is it you don't have to worry about the plus constant
on the left hand side we can just read this down this is equal to and the integral of Cossacks is
past the signs okay that's write this down and let me just write down plus one right here and
let me emphasize that the class t1 it's after the sine X so let me use the parentheses right
here to close it to emphasize that okay now we are pretty much done with the integration part
with the solving part of the differential equation and because we're given this initial value in this
case I would like to plug in X is PI and that will give us Y will be 0 I would like to comply
to this values right here to solve a see it's because if you isolate the Y first and I want
to stop a cdata you wouldn't know if she should be positive or negative let me just do it the way on
to show you guys right here I want to solve the C now and I will show you guys what I mean by we
wouldn't know if C is negative or not they turn okay plugging 0 into this Y here we have C Rose
here we get a red color 0 and then plus 1 and then that sings up the square root and this is equal to
we have 2 sine X is PI and we close that and we add us see well technically we're just using the
C right here so you don't need to label to see one just a see okay 0 plus 1 is 1 it's good of
1 is just 1 this is equal to sine of pi is 0 and then C so of course you see C is equal to
1 and you're pretty much done so here you have square root of y plus 1 it's equal to sine of X
see in this case as you can see it's partly one right so we add a one after that at the
end this is number two - so we'll do it we would like to isolate y so let's go ahead square four
sides they cancel and we will have y plus one it's equal to sine of X plus one inside of this
big parenthesis and square that and of course we have to just subtract 1 on both sides and we see
that Y is equal to sine of X and then plus 1 and a squared and then minus 1 here is the answer ok
all right so real quick imagine if you isolate the Y first and this is where the C was right so
if you isolate the Y first you have why it's equal to sine of X plus the C which we don't know what
it is and you square that this is my sine of X and then minus one after that imagine you if
you get this first and you plug in this values 0 into y and this is going to give you this is sine
of pi for the X and you add a C after that and you screw it up and the minus one you see this
is zero and sine of pi is zero plus C so it's just seen inside we have C squared this is equal to C
squared and we have the minus one you can add one on both sides you have C squared is equal to one
you see if you isolate the Y first earlier you end up with C squared equal to one and are you
going to say C is 1 or C is equal to negative one well to decide you actually have to look
back to this part of the equation so you have to know that right here Z has to be positive one so
this right here it wasn't preferred that's why I didn't show you guys this right here a quick
summary is that whenever you can solve for C easily you go ahead and do it and that's why I
did right here I saw what I see right away and then I isolate the Y after that and this is it