In this introductory course on Ordinary Differential Equations, we first provide basic terminologies on the theory of differential equations and then proceed to methods of solving various types of ordinary differential equations. We handle first order differential equations and then second order linear differential equations. We also discuss some related concrete mathematical modeling problems, which can be handled by the methods introduced in this course. The lecture is self contained. However, if necessary, you may consult any introductory level text on ordinary differential equations. For example, "Elementary Differential Equations and Boundary Value Problems by W. E. Boyce and R. C. DiPrima from John Wiley & Sons" is a good source for further study on the subject. The course is mainly delivered through video lectures. At the end of each module, there will be a quiz consisting of several problems related to the lecture of the week.
3-11 Clairaut's Equation
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From the course by Korea Advanced Institute of Science and Technology
Introduction to Ordinary Differential Equations
69 ratings
Korea Advanced Institute of Science and Technology
69 ratings
From the lesson
FIRST ORDER DEFERENTIAL EQUATION 2
Meet the Instructors
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
Okay. As a last example,
I'd like to introduce to you the Clairaut's equation.
Clairaut's equation is the first order differential equation
of the form equation nine say y=xy' + f(y')
with the function f(t) is twice differentiable,
and second derivative is never vanishing.
Then differentiating the equation nine one more time.
That gives you the equation of y"(x + f'(y'))=0.
Let's confirm it.
So we are starting from y= xy' + f(y')
differentiate both sides then from the left, you get y'.
From the right, first by the product rule,
y'+ xy"+f'(y')y' and by the chain rule times y".
Okay?
This y on the left hand side.
and this is y' on right hand side that they cancelled out so you get 0= xy"+f(y')y".
Factorize this common.
Factorize double prime out and you will get y"(x+f'(y') and that is equal to zero, right?
So, that's the equation right here, okay?
So that we have two possibilities, okay?
First, the y" can be equal to zero or x+f'(y') is equal to zero.
I will take the first chance, y"=0.
In other words,
y' is equal to some constant c.
Plugging that into the equation number nine then,
you will get y = cx + f(c) and this is the solution for arbitrary constant c.
That's easy to check, right?
From the same equation,
as you see what is y'?
That is equals C, right?
So plugging the y' is equal to c as a then you get c(x+f(c)), right?
So, trivially,
this family of straight lines is a solution to the given Clairaut's differential equation.
And since the Clairaut's equation is of the order one and this,
the family, has a one arbitrary parameter.
The family y=c(x+f(c)) is already a general solution
of this Clairaut's differential equation, okay?
And please note that by the form of this solution
you can recognize that the solution is a straight line depending on the slopes c.
So we get a family of solutions.
One parameter family of a solutions all over which are straight lines in the plane, okay?
Moreover,
the given Clairaut's differential equation nine has a one more solution,
which is a singular solution given by the parametric form say,
x = -f'(t), and y= f(t) - tf'(t).
That's my claim.
This parametric form of the equation is
also a solution of the given Clairaut's equation number nine.
Let's confirm first, okay?
To confirm it, let's compute y'.
By y' I mean dy/dx, okay?
That, by the chain rule, that is equal to the dy/dt(dt/dx), right?
So that is equal to dy/dt(dx/dt),
an inverse way. What is dy/dt?
From this equation down there,
dy/dt is easy to compute, right?
So we have y=f(t) - t(f'(t)).
Let's compute from this dy/dt, right?
That is the f'(t) and minus,
by the product rule.
This is f'(t) - f"(t), right?
So this two canceled out, right?
So simply, we get dy/dt,
that is the -tf"(t), right?
That we have down there, -tf"(t).
On the other hand, what is dx/dt?
dx/dt from this expression,
that is simply -f"(t), right?
And we need this the inverse way and multiply these two,
you simply get y'=t.
When y'=t, the right hand side of the Clairaut's differential equation,
which is the xy' + f(y') becomes, right?
What is xy'?
y' is equal to t we know,
and then what is x?
That is -f'(t) so the first term becomes a -tf'(t), right?
And plus f(y'), since the y' is t,
that is equal to f(t), right?
What is this one?
-f'(t)t + f(t) by this second equation down there,
that is exactly y.
And the y= xy' + f(y').
This is the original Clairaut's differential equation.
That simply means that the equation given by
this parametric form is really a solution to this problem,
the Clairaut's differential equation and this is the
singular because you can see very easily that
the function given by this parametric form cannot be a straight line and remind you that
the family of the general solutions of
the given Clairaut's differential equation is
a family of straight lines given by this equation,
y= cx + f(c), right?
Because of this,
another solution we just obtained is not the equation of the straight line.
It cannot be obtained from the general solution, and that means,
this solution is a singular solution, okay?
Geometrically,
this singular solution is nothing but the envelope of this family of straight lines, okay?
What is the envelop of something?
They are just simply the cover whose tangent lines are given by the family.
So, this over the with the envelope
over the family with straight lines is given by y= cx = f(c).
We will confirm it through the very simple example, okay?
So let's consider the following example.
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