00:00so now we'll discuss about the separable equation so in a separable equation like first is the it
00:13is the first order order differential equation which can be written in a way where y's terms
00:19are on one side and x terms are on x side so now taking the example let's say dy by dx is
00:28equal to x by y so let's like cross multiply this so y into dy y is equal to x into dx so as we can see
00:42that the x terms are on one side and y terms are on one side so this is one of the separable equation
00:50so let's take another example so let's say dy by dx is equal to
01:00uh x plus five one uh y so same way like you can also uh do this
01:21like dy by dx is equal to f of x into g of y so like we can divide by this and we'll double this
01:33so it will be d of y by g of y is equal to f of x into d of x so uh such type of equation where x
01:47term can be separated on one side and y terms can be separated on other side is known as the
01:53separable equation now talking about the next topic which is the integrating factor method
02:04so now talking about this so in integrating factor factor method it is a first order linear equation
02:12equation like for the first order linear ordinary differential equation uh this is the form uh y dash
02:19plus py is equal to q and so now talking about how like we have to solve this like in this uh
02:33we cannot separate x on the one x one side and y on the other side so there are two methods to solve
02:42this talking about the first method which is a formula method so uh in the formula method like what uh we'll do is that
03:02we'll just uh apply the formula uh to solve this uh equation so now talking about what the formula is
03:18so it will be like uh y is equal to e raised power minus p into dx and then we multiply by
03:32q dot e raised power p dx dx plus c so like first we will solve the p dx so let's first also here uh so now let's take an example
03:49now uh take example y dash plus uh
03:55uh three y is equal to uh uh three y is equal to uh five so now first solving through this equation we'll solve
04:12uh e raised power three into dx will be our e raised power 3x and now uh once we have this value we will put
04:22uh in this uh in this uh equation so y is equal to is power minus 3x and the bracket uh q the value of q is here
04:325 dot is power 3x into dx plus c
04:39now solving this y is equal to e raised power minus 3x and uh it will be 5 by 3 e raised power 3x
04:47plus c so uh this uh is the final answer like we can uh we can uh transfer this 3s power 3x here
05:06and uh yeah so now talking about the second method so this was the formula method
05:17and now talking about the second method
05:28which is by the procedure
05:36so the equation was y plus 3y plus uh is equal to 5
05:42so the first uh what we will do is that we will do uh find integrating factor
06:00and to find uh this what we will do is that uh e raised power
06:06is equal to 3a into dx uh so the value of integrating factor will be e raised power 3x
06:15now second multiply uh the differential equation by integrating factor
06:26multiply this uh equation by
06:30uh integrating factor so what will be uh y dash e raised power 3x is equal to uh plus 3y e raised power 3y
06:47is equal to 5 e raised power 3x and now what we will do we will integrate so while integrating
06:57uh so here it will be uh so here it will be y e raised power
07:073x because uh firstly our differentiator y dash it is for 3x and now it's a division in the second part
07:15so it will be y it is power uh this three will come so 3x so this is the derivative of
07:23d y by y y is power 3x so this is the derivative of this and here uh uh uh
07:31derivating the uh integrating this
07:40will be 5 by 3 e raised power 3x plus c so as you can see that we are getting the same equation
07:48y y is equal to 5 by 3
07:59plus
08:03c e raised power minus 3x and what was this
08:08so yeah so yeah this is the same so we are getting the same value by the second method also
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