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(a) What can you say about a solution of the equation $y^{\prime}=-y^{2}$ just by looking at the differential equation?

(b) Verify that all members of the family $y=1 /(x+C)$ are solutions of the equation in part (a).

(c) Can you think of a solution of the differential equation $y^{\prime}=-y^{2}$ that is not a member of the family in part (b)?

(d) Find a solution of the initial-value problem

$$y^{\prime}=-y^{2} \quad y(0)=0.5$$

$$

y=\frac{1}{x+2}

$$

Differential Equations

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Numerade Educator

Campbell University

Harvey Mudd College

University of Nottingham

Hello. Let's start with the first part of the question. What can what can we still say about the differential equation y prime equal to negative y square. So over here in the left hand side, we have the derivative of a function. And in the red inside we see that we have wife square that is always positive or zero. And with negative. This means that why prime is gonna be always less or equal to zero. So a derivative of a function, uh, negative or zero means that our function is going to be decreasing over. Um, always is gonna be decreasing on that interval, or we are gonna have. So if it is zero, then we might have stashed in our appointments a one for that function. So the second part of the question is I have to verify that why equal to one over X plus e is actually a family oak solutions for these difference regulation given here. Um, so basically, let's start with the left hand side. Eso we're We're going to get what prime equal to one over x plus e. Take the derivative of this. So we're going to get X plus C negative one derivative. So by definition, off the derivative, we're going to get negative one and then one over X plus C square times X plus C derivative. So, in other words, this is gonna be negative. One over eggs last see everything square because the derivative off X plus e is one. And on the on the other side we have native y square, so we have to prove or to see what these gets us. So this kids, this one over x plus e everything square which actually we have inequality over here. So the left inside and the right inside are the same. So this means that why equal to one over X Plessy is actually family of solutions for the differential equation given in one. The formula that I'm using over here is Theis. So in the general case, if we have a function f of x toe, some power and and we want to find the derivative of that so this is going to be in f of X at power and minus one times the derivative off the function. Okay, So in our case, we had explicitly the dirt do of that is one. Well, let's start with the other part. The other part is if we can think about other so solutions other than, um solutions off. Why I go to native Why square other than those given in the family of social. So I'm thinking about why I called zero. It satisfies the condition. So why prime here? So we have What is your prime equal to negative zero square. So we have zero equal to zero and we're good, but this is not part off the family of solutions. Why? Over why you could one over X plus e Because there is, there does not, um, exist Any exercise e such daddy, we if we're substituted, don't Over here we get why Equals is here in the last part is actually related toe find the solution for white prime equal to negative while square where it's given why it zero is 0.5. There are two ways. Actually, we could eso the differential equation here and find family of solutions and then find a particular solution that satisfies this initial condition. Or we can use the family off solutions that we proved on under to that it was a family of socials for this depression equation and then find a particular solution for with for these initial conditions. So I'm using the family off solutions, which which is one over x plus e. And we know that, um, so we know that. Why at we know that, why at zero, which is your A 00.5 is gonna give us 1/0 plus a constant. So from here will gather the constant physical toe to and we get the the dealer solution, which is why equal toe one over eggs plus toe. Okay, so this is it. So this is one. The solution that we are looking for that satisfies the initial condition. So these are all the solutions for this question. He and I have to get more questions to answer Morse to give more solutions and make more videos. Thank you.