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One more example.

Â Now, let's think about the initial value problem,

Â differential equation is a 6xydx + (4y + 9x squared) dy = 0.

Â And initial condition is y(0) = 1.

Â Again, let's check the differential equation,

Â to be exact or not.

Â dM over dy - dN over dx.

Â From this one, dN over dy that is a 6x.

Â From this expression, dN over dx is 18x.

Â Subtract it, you get negative 12 times x which is not identical as 0.

Â So, the given equation is not exact.

Â Let's compute the following quantity.

Â 1 over M (dM over dx minus dM over dy).

Â Then that is equal to 2 over y

Â because this quantity is negative 12x divided by 6xy that is equal to 2 over y.

Â So, this is a function of y only.

Â Then there is an integrating factor on view which is also a function of a y only.

Â And in fact, v over y is a given by

Â exponential integral of 2 over y_dy that is equal to y squared.

Â So this is an integrating factor.

Â Multiply the given equation by this integrating factor.

Â So you get 6xy cubed dx +

Â (4y cube + 9x squared y squared) dy is equal to 0.

Â And this must be exact.

Â This must be exact. What does that mean?

Â Again, there must be a function capital F(x,y)

Â whose x partial is equal to 6xy cubed and

Â whose y partial derivatives is 4y cubed + 9x squared y squared.

Â Let me use the first equation.

Â Say, dF over dx is equal to 6xy cubed.

Â Then capital F(x, y) through the x integration it'll be

Â 3x squared y cube plus arbitrary integral constant which is arbitrary function of,

Â arbitrary differentiable function of y, g(y).

Â Now take y partial of this capital F. You have a 9x squared y squared

Â + g prime of (y) must be equal to, dN.

Â dN, right here.

Â dN, right here.

Â So, compare these two equations then.

Â 9x squared y squared cancelled out,

Â and then this g prime of (y) is equal to to 4y cubed.

Â So then this g(y) must be equal to y to_the_four.

Â Therefore, 3x squared y cubed + y

Â to_the_four is equal to arbitrary constant c. That's a general solution.

Â Finally, let's use the initial condition y(0) = 1.

Â When x is equal to zero,

Â y is equal to one,

Â that gives immediately c is equal to one.

Â So the solution of the initial value of problem is

Â 3x squared times y cubed + y to_the_four.

Â And that is equal to 1.

Â So far, I introduced

Â a subclass of first order differential equations which are rather easy to solve.

Â For example, I introduced the linear first order differential equation,

Â and the separable first order differential equation,

Â and the exact differential equation,

Â for which we have a simple analytic tools to solve those differential equations.

Â But there are many other- the first order differential equation,

Â which is neither linear,

Â separable or nor exact.

Â In this section, I would like to show

Â some other subclass of first order differential equations which may be

Â transformed into another form by a suitable substitution,

Â which is easy to solve.

Â As a first of such example,

Â let's consider this as specific differential equations.

Â Say, xe to_the_y y prime + e to_the_y = cos x.

Â Just look at this differential equation carefully.

Â And then, try the following.

Â I will set u = e to_the_y.

Â Then, u prime is equal to by the chain rule,

Â derivative of exponential function is itself.

Â And because of y is the function of x,

Â so derivative of e to_the_y will be e to_the_y times y prime.

Â With this a simple substitution,

Â the given differential equation becomes now e to_the_y y prime,

Â there is a u prime.

Â So xu prime + e to_the_y, that is u,

Â and the xu prime + u is the same as derivative of x times u.

Â Then, it must be equal to cosine of x.

Â Now, look at the differential equation we obtained now for the due unknown function u.

Â This is the first order differential equation,

Â which is very easy to solve.

Â What is x times u of x?

Â It's sine to derivative of cos x.

Â In other words, that is sin x + c. So now, we have this one.

Â x times u, that is x times e to_the_y because u is equal to e to_the_y,

Â that is equal to sin x + c. That's a general solution of the given differential equation.

Â We can solve this first to the differential equation through a very simple substitution.

Â Say, u is equal to exponential y.

Â