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.