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As a second class of dispersion first order, differential equation,

Â let me introduce to you the following.

Â A first order differential equation is linear if it is of the form

Â a sub 1 of x times the y prime plus a naught of x times y is equal to g of x.

Â So really linear in the variables of y and this derivative way.

Â So we call it as a linear first order differential equation.

Â Furthermore, if this right hand side,

Â the g of x is identical to zero,

Â then we call it as homogeneous first order linear differential equation.

Â Otherwise, in other words,

Â if g of x is not identically equal to zero then we record it as

Â nonhomogeneous first order linear differential equation.

Â If right hand side is identical to zero then homogeneous.

Â Otherwise, we record it as a nonhomogeneous.

Â Let us say, a naught

Â over a 1 to the p of x and g over a sub 1 to be q of x.

Â Then starting from the original equation

Â which I reminded here, differential equation one,

Â it was a sub 1 of x y prime

Â plus a naught of x y and that is equal to g of x, right.

Â Where people call this a sub 1 of x has a leading coefficients.

Â This is the coefficient of the highest derivative of y appearing in this expression way.

Â So divided through a 1 of x divided through the equation through the a 1 of x,

Â then you are going to get y prime plus a naught over

Â a 1 times y that is equal to g over a sub 1.

Â We call this a second equation as an domain form of the original equation.

Â I introduce just another notation for this part,

Â I will call it to be letter p. This part I will call it to be letter q.

Â So we have expressions to y prime plus p of x times the y is equal to q of x.

Â This is a standard form of the linear first order differential equation.

Â Very special case if p x is equal to identical zero

Â then the equation reduce to y prime is equal to q of x.

Â This is the simplest the differential equation

Â whose general solution is very simply given by general solution y is equal

Â to any anti derivative of q which I denoted by integral q of x of

Â the x plus arbitrary integral constant to c. This is a general solution,

Â as we know quite well from the integral calculus.

Â In general or in other words the if p of x is not identically zero,

Â what can we do then?

Â In that case what we are going to do is the following thing.

Â Okay. Here we have a standard form of the first order linear differential equation.

Â Let us say y prime plus p of x times the y and that is equal to q of x.

Â This is a so called standard form of the first order linear differential equation.

Â I am looking for some unknown function Âµ of x,

Â Âµ of x such that if I multiply this equation by

Â Âµ then the right hand side

Â will be Âµ times the q a.

Â And the other hand,

Â of the left inside will be Âµ y prime plus Âµ p times y. I like to choose Âµ,

Â this unknown function Âµ.

Â So that, this right hand side is the same as

Â Âµ times y and this derivative way.

Â I would like to choose a Âµ so that this equality holds.

Â What does that mean?

Â Just divide a product for the derivative of Âµ times y is the same

Â as Âµ prime y plus Âµ y prime by the product through, right.

Â Again, let us set up the equality between this and that expression.

Â Then here you have a Âµ y prime.

Â Here you have the same term down there so that we can cancel them out.

Â Then finally you are going to get Âµ prime of y is equal to Âµ over p times of y.

Â In which y is common again so that

Â you may require that Âµ prime is equal to p times of Âµ.

Â So what I am looking for is,

Â I am looking for some unknown function Âµ,

Â such that if I multiply the given first order differential equation by this Âµ.

Â Then Âµ times the y prime plus

Â Âµ times the p of y can be written as a derivative of Âµ times of y.

Â Which is the same as the Âµ prime is equal to p times of Âµ.

Â So we get another first order differential equation.

Â But now, this differential equation is a separable, right.

Â Can you see it?

Â We have two variables,

Â one independent variable x,

Â and the unknown function Âµ,

Â this is an other dependent variable.

Â So let us separate the variable then you are going to get one over Âµ

Â and d Âµ that is equal to p of x and the d of x.

Â We really separate the variables here.

Â How to solve Âµ? Take intuition of both sides.

Â From this side you are going to get log of absolute value of a Âµ.

Â From this side anti derivative of letter p of x, d of x.

Â Any one of the anti derivative of it plus arbitrary constant of c,

Â c sub 1 for example.

Â You are going to get this expression.

Â So, for the absolute value of the Âµ will be

Â equal to e to the c 1 into letter p of x and the d of x.

Â So, this is my mistake because I am taking the exponential.

Â So, it should be e to the anti derivative into letter p x and d of x.

Â We did it this type of thing the couple times already. What is a Âµ?

Â Âµ is equal to plus or minus into this c 1,

Â e to the p of x and the d of x to get this work.

Â So what does that mean?

Â If I choose Âµ of x unknown function Âµ of x to be,

Â this expression down there.

Â If I multiply the given differential equation by any one of such functions in

Â Âµ then the problem becomes Âµ times the q.

Â Now Âµ is given by this one that is equal to derivative of Âµ of y.

Â That is a very one of the simplest differential equation that we can solve.

Â So here, you have some arbitrary choice of c 1,

Â but that you may call it as c.

Â Now the way the c is equal to plus or minus e to the c

Â 1 which is the arbitrary one non zero real number.

Â So, why not choose it?

Â Because we do not need all sorts of possibilities for Âµ.

Â We need the only one possibility.

Â So, why not choose the c is equal to one?

Â That is my choice down there.

Â Why not choose c is equal to one?

Â By choosing c is equal to one,

Â you will get Âµ of x is equal to exponential integral p x and d x.

Â That is the equation for down here.

Â So, multiply the given standard form of

Â the first order linear differential equation by this function Âµ of x.

Â Then the problem becomes a very simple one which is easy to solve.

Â As you can see now.

Â So, multiply with this Âµ with choice of c is equal to

Â one so multiply this whole equation by this Âµ then.

Â Then left hand side will be Âµ times

Â the q and the right hand side will be derivative of Âµ times y.

Â So you are going to get this equation.

Â And that means what?

Â You can not solve this equation very easily.

Â What is the Âµ x of y?

Â Âµ x of y must be equal to

Â anti derivative of Âµ x times the q of x.

Â Any one of this said derivative plus arbitrary constant c. Now,

Â from this expression divided the whole thing by the Âµ of x,

Â you are going to get y is equal to 1 over

Â Âµ x times the integral Âµ x times the q x and d x plus

Â arbitrary constant c. This is the general solution

Â of our given first order linear differential equation down there.

Â So, once we find the Âµ of x which is given by this expression then we can

Â find the general solution of this linear

Â first order differential equation in the form of the equation five.

Â We call this a specific choice of Âµ,

Â Âµ is equal to exponential integral p x of d x

Â as an integrating factor for the linear differential equation, right.

Â For this linear differential equation we call

Â this the Âµ of the x to be on integrating factor, okay?

Â