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Now, let's introduce another one,

Â which is the so-called Bernoulli equation,

Â a first order differential equation of the form,

Â y prime plus pxy is equal to qx times y to the n,

Â n is a constant, okay?

Â Any real constant.

Â It's called the Bernoulli equation.

Â Here, I have shown n is not equal to 0 or 1,

Â and qx is not identical to 0.

Â Why? If otherwise, if n is equal to 0,

Â then it is like a y prime plus pxy is equal to q of x.

Â That's a linear equation.

Â If n is equal to 1, y prime plus pxy is equal to qxy.

Â That's, again, a linear equation, okay.

Â If qx is identical is 0, then it's simpler,

Â y prime plus pxy is equal to 0, right?

Â In all those cases,

Â the problem is a linear problem,

Â which we can handle easily so that we assume that n is not equal

Â to 0 or 1 or qx is not identically 0, okay?

Â Note here that before giving the general solution of this Bernoulli equation,

Â okay, also have the following effect.

Â If n is any positive constant,

Â if n is any positive constant,

Â then y is equal to identically zero,

Â it's the trivial solution.

Â Can you see that? If this power is a strictly positive,

Â and if y is equal to identical is zero,

Â then right hand side is zero and left hand side is the zero, right?

Â So, the differential equation is trivially satisfied.

Â So, y is equal to 0, constant function.

Â This is the trivial solution, okay?

Â Now, let's look at this equation again, Bernoulli equation,

Â divide this equation, divide the both sides by y to the n,

Â then you are going to get,

Â y to the negative n y prime plus p

Â of x y to the 1 minus n and that is equal to q of x, right?

Â Divide through the divide of the given Bernoulli equation by y to the n,

Â then you are going to get this one, okay?

Â Now set, u is equal to y to the 1 minus n, okay? My claim is that this is a substitution.

Â Transform this Bernoulli equation into a linear equation say,

Â u prime plus 1 minus n times pxu is equal to 1 minus n q of x,

Â which is a really linear force

Â to the differential equation for the unknown function u.

Â Why is it that? It's simple, right?

Â So, from this equation I set, u is equal to y to the 1 minus n, right?

Â What happened then? What is u prime?

Â u prime is equal to 1 minus n times Y to the negative n times y prime, okay?

Â Look at this part.

Â Look at this equation,

Â y to the negative ny prime. That is, 1 over 1 minus nu prime , right?

Â So, the left hand side,

Â it becomes 1 over 1 minus nu prime plus px,

Â times what is y to the n minus n?

Â That is u, and that is equal to q, right?

Â So multiply through by 1 minus n,

Â you are going to get

Â this linear force to the differential equation for the unknown u, right?

Â That is my claim, okay?

Â And be careful, here,

Â I'm dividing the equation through by 1 minus n, right?

Â Since I'm assuming that n is not equal to one,

Â 1 minus n is never 0 so that we can safely divide this the question by 1 minus n, right?

Â That's another reason why I'm assuming that n is not equal to one, okay?

Â Anyway, this last first order linear differential equation,

Â we know how to handle it, right?

Â That means we know how to handle this Bernoulli equation, okay?

Â I will check this thing through the exam first.

Â The problem is so y prime plus 2 over xy is equal to 2x times y to the 1/2, right?

Â What is n in this example?

Â n is equal to one half, right?

Â Which is not equal to zero,

Â not equal to one, okay,

Â divide the equation through y to the 1/2, right?

Â Then, you are going to get by y to

Â the negative 1/2y prime plus 2 of xy to the 1/2, and that is equal to 2 of x, right?

Â Set y is equal to y to the 1/2, right?

Â Set u is equal to y to the 1/2, right?

Â Then u prime is equal to 1/2 times y to the negative 1/2 y prime.

Â So y to the negative 1/2y prime,

Â it's the same as the two times over u prime and this is equal to 2 over xn times u,

Â that is equal to 2x, right?

Â Divide it through by 2 then,

Â you are going to get u prime plus x to the negative 1 times u, that is x.

Â Multiply through by x then,

Â you will get xu prime plus u,

Â which is the same as derivative of x times u,

Â that must be equal to x squared, right?

Â