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.