how to solve separable differential equations
welcome a video series in differential equations and will first talk about the first kind of
differential equations that we have to learn how to solve these separable differential equations
in this video we'll first talk about the idea the general strategy and also show us one example of
how to stop a differential equation by separating the variables and you can watch my other videos I
will have more examples including home solutions for you guys alright this is how we should be get
notice that we have to isolate that dy/dx first right and you see this is just a first derivative
therefore this is also considered to be the first order differential equation sometimes this can
be written as Y Prime in that case just ready dumbass dy over DX maybe we are using T for the
independent variable then dy DT then things like that alright let me write this down again dy DX
our goal is to focus on the right-hand side first this right here is a function in terms of both x
and y together our first job is to separate the X and the y by factoring look at the right-hand
side I want of battery as a function and I was just ready done G of X since pressure in terms
of just X and the second part is going to be multiplying with H of Y a function in terms of
just Y and this is what we mean by start reading the variable next and I will suggest you guys to
do it this way let's multiply the x on both sides whenever we're trying to solve a differential
equation that differential should not be in the denominator so let's take care of that first and
then we will have dy equals to G of x times H of Y and then we have the DX right here and then what
you want to do is you want to have all the Y's to get on one side and the etiquette on one side you
see we have h of wild rice here that should be together with the y so that the pipe o set by H
of Y so they cancel divided by H of Y as well and that G of X is together with DX so that's great
on the left hand side let me write it down as one over H of Y and then let's put a DUI on the side
and this is equal to G of X yet and you see this is warming by all the Y's together and all the
eggs together and congratulations to solve this kind of differential equation all we have to do
this integrate this integrate that so we just have to do the integrals twice okay and hopefully this
is napa to integrate likewise this is not bad to integrate as well and now let's take a look
of an example and let's take a look at how we can solve this differential equation we have dy/dx is
equal to x plus x times y squared as I mentioned the earlier the strategies have to dy DX on one
side which we do right so now let's focus on the right hand side can we factor things out yes we
can factor out the X because both terms have the X right so let me try this down again dy/dx this
is going to be that's factor the X so we have X in the front and we have 1 plus y squared like this
and now I want to show guests all the steps let's multiply the x on both sides this way the DX will
cancel and what this right and we have to divide the 1 plus y squared on both sides so we can bring
this to the other side so let's do that divide this by 1 plus y squared and right here let me
read it down as 1 over 1 plus y squared and then you see this will cancel on the left hand side
now we'll just have 1 over 1 plus y squared and that's put a DUI on the side and then we will have
this equal to X with the DX right here and this is forming by this step okay and now I'm just gonna
integrate this integrate that and once again I'm gonna show you guys all the steps so let's focus
on this part here first the integral of 1 over 1 plus y squared in the wide world it's going
to be the inverse tangent of Y and technically you wanna integrate this I will have to add a
constant and I will add a constant right here but let me put this down as plus c1 all right
let's focus on the right hand side this is equal to when we integrate X in the X world we have 1/2
x squared and this is the integral we should also add a constant at the end right so let me add plus
C but this time let me see - but as I said I'm just showing case all the steps in fact you can
skip this adding c1 a DC - step because what you want to do next is minus c1 on both sides so that
we can so and you see you have the inverse tangent Y and this is equal to 1/2 X square and now this
is the fun part here we have positive C 2 and then minus D 1 a constant minus another constant
is just another constant so I can label this s just seems three so i can register it plus c 3
and c 3 is going to be C 2 minus C 1 okay but as I said later I can just go from here to here and
then begin to label your constant right here's one but anyways this is how I will show you right now
well this is not bad to isolate the Y so we will do that that's isolate the Y so to do that we'll
have to take the original tangent on both sides so that tangent and the intentional cancel our
do that right here as well we take the original tangent be sure you put both the 1/2 x squared and
Plus these three both in South apprentices at the end you will have why it's equal to the original
tangent of 1/2 x squared plus c3 and we're pretty much done right so let me just put on a seat as a
legitimate see that want to use that yet sometimes some books let me use the capital K right here for
the final constant up to that right but this right here is yet this is holding by a general
solution for this differential equation and this is it however I want to show it gets more because
suppose you're given an initial value let's say you have this differential equation along with
an initial condition let's say Y of 0 this is equal to negative 1 and you want to stop this
initial value problem and this is how you have to do it solve this this way we get a general
solution right and our goal is to figure out what this series we have to use this initial condition
and the way that this is written it means that when you have this through inside this means X
is equal to 0 and Y is equal to negative 1 Y it's a function of X that's why this 0 means
X is equal to 0 and the output means Y is equal to negative 1 well what we do next look at the
general form plugging negative 1 into this wife here and this is equal to tangent of 1/2 and the
X is equal to 0 so we're plugging the symbol right here and we'll square that and I'm a plus thing
like this right all right so this is going to be negative 1 equals 2 tangent this is just 0 so we
just have a see inside so negative 1 is equal to tangent C well tangent of 1 and go what gives you
negative 1 we know tangent of negative PI over 4 would give you negative 1 right therefore C is
equal to negative PI over 4 so with this initial condition you'll be able to find out what's this
which is negative power for in this situation at the end just liking right here finally you
can say why it's equal to tangent of 1/2 x squared C is negative PI over 4 so I just have to plug in
right here so let me say minus PI over 4 like that so this is House of the initial value problem and
once again I will have more videos for more examples and home or solutions for you guys