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Okay. As a last example,

Â I'd like to introduce to you the Clairaut's equation.

Â Clairaut's equation is the first order differential equation

Â of the form equation nine say y=xy' + f(y')

Â with the function f(t) is twice differentiable,

Â and second derivative is never vanishing.

Â Then differentiating the equation nine one more time.

Â That gives you the equation of y"(x + f'(y'))=0.

Â Let's confirm it.

Â So we are starting from y= xy' + f(y')

Â differentiate both sides then from the left, you get y'.

Â From the right, first by the product rule,

Â y'+ xy"+f'(y')y' and by the chain rule times y".

Â Okay?

Â This y on the left hand side.

Â and this is y' on right hand side that they cancelled out so you get 0= xy"+f(y')y".

Â Factorize this common.

Â Factorize double prime out and you will get y"(x+f'(y') and that is equal to zero, right?

Â So, that's the equation right here, okay?

Â So that we have two possibilities, okay?

Â First, the y" can be equal to zero or x+f'(y') is equal to zero.

Â I will take the first chance, y"=0.

Â In other words,

Â y' is equal to some constant c.

Â Plugging that into the equation number nine then,

Â you will get y = cx + f(c) and this is the solution for arbitrary constant c.

Â That's easy to check, right?

Â From the same equation,

Â as you see what is y'?

Â That is equals C, right?

Â So plugging the y' is equal to c as a then you get c(x+f(c)), right?

Â So, trivially,

Â this family of straight lines is a solution to the given Clairaut's differential equation.

Â And since the Clairaut's equation is of the order one and this,

Â the family, has a one arbitrary parameter.

Â The family y=c(x+f(c)) is already a general solution

Â of this Clairaut's differential equation, okay?

Â And please note that by the form of this solution

Â you can recognize that the solution is a straight line depending on the slopes c.

Â So we get a family of solutions.

Â One parameter family of a solutions all over which are straight lines in the plane, okay?

Â Moreover,

Â the given Clairaut's differential equation nine has a one more solution,

Â which is a singular solution given by the parametric form say,

Â x = -f'(t), and y= f(t) - tf'(t).

Â That's my claim.

Â This parametric form of the equation is

Â also a solution of the given Clairaut's equation number nine.

Â Let's confirm first, okay?

Â To confirm it, let's compute y'.

Â By y' I mean dy/dx, okay?

Â That, by the chain rule, that is equal to the dy/dt(dt/dx), right?

Â So that is equal to dy/dt(dx/dt),

Â an inverse way. What is dy/dt?

Â From this equation down there,

Â dy/dt is easy to compute, right?

Â So we have y=f(t) - t(f'(t)).

Â Let's compute from this dy/dt, right?

Â That is the f'(t) and minus,

Â by the product rule.

Â This is f'(t) - f"(t), right?

Â So this two canceled out, right?

Â So simply, we get dy/dt,

Â that is the -tf"(t), right?

Â That we have down there, -tf"(t).

Â On the other hand, what is dx/dt?

Â dx/dt from this expression,

Â that is simply -f"(t), right?

Â And we need this the inverse way and multiply these two,

Â you simply get y'=t.

Â When y'=t, the right hand side of the Clairaut's differential equation,

Â which is the xy' + f(y') becomes, right?

Â What is xy'?

Â y' is equal to t we know,

Â and then what is x?

Â That is -f'(t) so the first term becomes a -tf'(t), right?

Â And plus f(y'), since the y' is t,

Â that is equal to f(t), right?

Â What is this one?

Â -f'(t)t + f(t) by this second equation down there,

Â that is exactly y.

Â And the y= xy' + f(y').

Â This is the original Clairaut's differential equation.

Â That simply means that the equation given by

Â this parametric form is really a solution to this problem,

Â the Clairaut's differential equation and this is the

Â singular because you can see very easily that

Â the function given by this parametric form cannot be a straight line and remind you that

Â the family of the general solutions of

Â the given Clairaut's differential equation is

Â a family of straight lines given by this equation,

Â y= cx + f(c), right?

Â Because of this,

Â another solution we just obtained is not the equation of the straight line.

Â It cannot be obtained from the general solution, and that means,

Â this solution is a singular solution, okay?

Â Geometrically,

Â this singular solution is nothing but the envelope of this family of straight lines, okay?

Â What is the envelop of something?

Â They are just simply the cover whose tangent lines are given by the family.

Â So, this over the with the envelope

Â over the family with straight lines is given by y= cx = f(c).

Â We will confirm it through the very simple example, okay?

Â So let's consider the following example.

Â