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So, we are required to solve first order nonlinear equation y is

Â equal to x plus four times y_prime and plus y_prime_squared.

Â First, we have to recognize it as the Clairaut's differential equation.

Â Look at this equation then.

Â Let me write it here.

Â y is equal to x plus four times y_prime plus y_prime_squared.

Â And do you remember the general form of Clairaut's differential equation?

Â It was the, yeah, right here,

Â y is equal to x times y_prime plus some function of y_prime.

Â So, let's rewrite this expression down there as x

Â y_prime plus four y_prime plus y_prime_squared.

Â This part must be f of y_prime.

Â So you can read it out.

Â This is x y_prime plus f of y_prime.

Â What is f? F of t is equal to, can you read it?

Â Simply four t plus t_squared.

Â That's what I write right here.

Â The given the equation down here,

Â we can recognize it as y is equal to x times y_prime plus f of

Â y_prime where f of t is equal to t_squared plus four t. So,

Â the given differential equation is Clairaut's differential equation.

Â So what is general solution?

Â It's easy enough, right?

Â Y_prime is equal to arbitrary constant c,

Â so that y is equal to cx plus f of c. F of c is c_squared plus four c. So,

Â y is equal to cx plus c_squared plus four times of C. This is the general solution,

Â family of straight line.

Â And we have one other extra solution which is a singular solution given in

Â the parametric form x is equal to

Â negative two t minus four and y is equal to negative t_squared.

Â Where did we get this one?

Â Let's go back to the general theory.

Â We have for the general Clairaut differential equation,

Â we have a singular solution given by x is equal to

Â negative f_prime of t. Let's remind it here, right?

Â X is equal to negative f_prime of t and y is equal to f of

Â t minus t times f_prime of t. Let's remind it here,

Â then apply it to our example.

Â In our example, what is f of t?

Â In our example, we have f of t is equal to t_squared plus four t, right?

Â T_squared plus four t. So I will just copy the things down there.

Â So, what was it?

Â X is equal to negative f_prime of t and y is equal

Â to f of t minus t times f_prime of t, right?

Â This is the parametric form of the solution where we have now

Â have f of t is equal to t_squared plus four t, right?

Â So plugging those two expressions then,

Â you can easily see that x is equal to negative f_prime,

Â the derivative of f is equal to two t plus four

Â and this negative so you get here x is equal to negative two t

Â minus four and y is equal to f of t minus t times f_prime

Â of t. Through this computation you will get y is equal to negative t squared.

Â In this example, you can eliminate the parameter t through this two equation because,

Â from the first, you have x is equal to negative two t minus four.

Â What does that mean?

Â That is equal to t is equal to x plus for over negative two,

Â and then y is equal to negative t_squared.

Â Negative t_squared, that is equal to negative x plus four squared over four.

Â That's the equation of the parabola given here by the red lines, red curve.

Â What I mean is exactly this part.

Â This part is an equation of the envelope,

Â y is equal to negative x plus four and squared over four.

Â That's the singular solution of the given Clairaut differential equation.

Â On the other hand,

Â the general solution which is a family of straight lines are given by y is equal to cx

Â plus c_squared plus four c. I draw several such solutions for different values of c,

Â where c is equal to zero.

Â When c is equal to zero,

Â y is equal to zero.

Â That means the x-axis, right here,

Â which is a tangent to the parabola.

Â When c is equal to one,

Â this straight line in blue lines,

Â this is also tangent to the parabola.

Â And when c is equal to negative two and when c is equal to negative one,

Â you always get a straight line which is tangent to this curve.

Â That's what I mean.

Â This parabola, y is equal to negative x plus four squared over four,

Â this is envelope of the family of straight lines given by

Â y is equal to cx plus c_squared and plus four c.

Â