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.
3-12 Clairaut's Equation - Ex. 7
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
157 個評分
Korea Advanced Institute of Science and Technology
157 個評分
從本節課中
FIRST ORDER DEFERENTIAL EQUATION 2
與講師見面
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
So, we are required to solve first order nonlinear equation y is
equal to x plus four times y_prime and plus y_prime_squared.
First, we have to recognize it as the Clairaut's differential equation.
Look at this equation then.
Let me write it here.
y is equal to x plus four times y_prime plus y_prime_squared.
And do you remember the general form of Clairaut's differential equation?
It was the, yeah, right here,
y is equal to x times y_prime plus some function of y_prime.
So, let's rewrite this expression down there as x
y_prime plus four y_prime plus y_prime_squared.
This part must be f of y_prime.
So you can read it out.
This is x y_prime plus f of y_prime.
What is f? F of t is equal to, can you read it?
Simply four t plus t_squared.
That's what I write right here.
The given the equation down here,
we can recognize it as y is equal to x times y_prime plus f of
y_prime where f of t is equal to t_squared plus four t. So,
the given differential equation is Clairaut's differential equation.
So what is general solution?
It's easy enough, right?
Y_prime is equal to arbitrary constant c,
so that y is equal to cx plus f of c. F of c is c_squared plus four c. So,
y is equal to cx plus c_squared plus four times of C. This is the general solution,
family of straight line.
And we have one other extra solution which is a singular solution given in
the parametric form x is equal to
negative two t minus four and y is equal to negative t_squared.
Where did we get this one?
Let's go back to the general theory.
We have for the general Clairaut differential equation,
we have a singular solution given by x is equal to
negative f_prime of t. Let's remind it here, right?
X is equal to negative f_prime of t and y is equal to f of
t minus t times f_prime of t. Let's remind it here,
then apply it to our example.
In our example, what is f of t?
In our example, we have f of t is equal to t_squared plus four t, right?
T_squared plus four t. So I will just copy the things down there.
So, what was it?
X is equal to negative f_prime of t and y is equal
to f of t minus t times f_prime of t, right?
This is the parametric form of the solution where we have now
have f of t is equal to t_squared plus four t, right?
So plugging those two expressions then,
you can easily see that x is equal to negative f_prime,
the derivative of f is equal to two t plus four
and this negative so you get here x is equal to negative two t
minus four and y is equal to f of t minus t times f_prime
of t. Through this computation you will get y is equal to negative t squared.
In this example, you can eliminate the parameter t through this two equation because,
from the first, you have x is equal to negative two t minus four.
What does that mean?
That is equal to t is equal to x plus for over negative two,
and then y is equal to negative t_squared.
Negative t_squared, that is equal to negative x plus four squared over four.
That's the equation of the parabola given here by the red lines, red curve.
What I mean is exactly this part.
This part is an equation of the envelope,
y is equal to negative x plus four and squared over four.
That's the singular solution of the given Clairaut differential equation.
On the other hand,
the general solution which is a family of straight lines are given by y is equal to cx
plus c_squared plus four c. I draw several such solutions for different values of c,
where c is equal to zero.
When c is equal to zero,
y is equal to zero.
That means the x-axis, right here,
which is a tangent to the parabola.
When c is equal to one,
this straight line in blue lines,
this is also tangent to the parabola.
And when c is equal to negative two and when c is equal to negative one,
you always get a straight line which is tangent to this curve.
That's what I mean.
This parabola, y is equal to negative x plus four squared over four,
this is envelope of the family of straight lines given by
y is equal to cx plus c_squared and plus four c.
