### Video Transcript

Consider the differential equation two dπ¦ by dπ₯ is equal to π₯ plus six π¦ minus 12. Part 1) On the axes provided, sketch a slope field for the given differential equation at the seven points indicated. Part 2) Find d two π¦ by dπ₯ squared in terms of π₯ and π¦. Describe the region in the π₯π¦-plane in which all the solution curves to the differential equation are concave down. Part 3) Let π¦ is equal to π of π₯ be a particular solution to the differential equation. If π of one is equal to two, does π have a relative minimum, a relative maximum, or neither at π₯ equals one? Part 4) Find the values of the constants π and π for which π¦ is equal to ππ₯ plus π is a solution to the differential equation.

For part 1, weβre required to sketch a slope field for the seven points indicated on the axes. These seven points are negative two, negative one; negative one, negative one; negative one, one; negative one, two; zero, one; one, two; and two, two. Now, letβs draw a table of values for these π₯, π¦ and corresponding dπ¦ by dπ₯ values. We know the π₯- and the π¦-values from the points given in the graph. Now, we just need to find the corresponding dπ¦ by dπ₯ values. And we can do this by substituting the values of π₯ and π¦ into our differential equation.

We can make this a little easier by dividing both sides of the equation by two. For the point negative two, negative one, we substitute in π₯ is equal to negative two and π¦ is equal to negative one. This gives us a value for dπ¦ by dπ₯ of negative 10. Next up, we have π₯ is equal to negative one and π¦ is equal to one. This gives us a value for dπ¦ by dπ₯ of negative 19 over two. Then, we have negative one, one giving us negative seven over two. We can continue like this for the rest of the points in our table. We find that the corresponding dπ¦ by dπ₯ values for the remaining points in the order that they appear in the table are negative one-half, negative three, one-half, and one.

Weβre now ready to sketch our slope field since dπ¦ by dπ₯ represents the slope at each of these points. We simply draw a line in with the slope of the corresponding dπ¦ by dπ₯ value for each of these points. And this will give us a slope field for the seven points. This is what our slope field will look like. And so, this completes part 1 of the question. Weβll now move on to part 2.

Weβre required to find d two π¦ by dπ₯ squared in terms of π₯ and π¦ and then describe the region in the π₯π¦-plane for which the solution curves of the differential equation are concave down. We have that two dπ¦ by dπ₯ is equal to π₯ plus six π¦ minus 12. In order to find d two π¦ by dπ₯ squared, we can use implicit differentiation. This means we simply differentiate both the left- and right-hand side of the equation with respect to π₯. And we can do this term by term.

On the left-hand side of the equation, we have two dπ¦ by dπ₯. When we differentiate this with respect to π₯, we obtain two d two π¦ by dπ₯ squared. Now moving to the right-hand side of the equation, our first term is π₯. When we differentiate this with respect to π₯, we simply get one. Our next term is six π¦. Differentiating this with respect to π₯ gives six dπ¦ by dπ₯. And our final term here is negative 12 which is just a constant. However, when we differentiate any constant with respect to π₯, we get zero. Now, we have an equation for d two π¦ by dπ₯ squared in terms of dπ¦ by dπ₯. Our other question asks us to find d two π¦ by dπ₯ squared in terms of π₯ and π¦.

We already have an equation for dπ¦ by dπ₯ in terms of π₯ and π¦. So we can substitute in this equation. It may be a little easier if we factor out two dπ¦ in our equation for d two π¦ by dπ₯ squared like this. Then, we can directly substitute the equation for two dπ¦ by dπ₯ into this equation. This gives us that two d two π¦ by dπ₯ squared is equal to one plus three multiplied by π₯ plus six π¦ minus 12. Next, we can expand the brackets to give that two d two π¦ by dπ₯ squared is equal to one plus three π₯ plus 18π¦ minus 36. And finally, we can simplify this by combining the one and the negative 36 and then dividing by two.

Now, we have found d two π¦ by dπ₯ squared in terms of π₯ and π¦. And we can rewrite it as follows: d two π¦ by dπ₯ squared is equal to three over two π₯ plus nine π¦ minus 35 over two. Next, weβre required to describe the region in the π₯π¦-plane in which all the solution curves for the differential equation are concave down. Next, we need to find the region in the π₯π¦-plane for which all the solution curves of the differential equation are concave down. Weβll be using the fact that when d two π¦ by dπ₯ squared is strictly less than zero, our curve is concave down. So we need to simply set d two π¦ by dπ₯ squared to be less than zero. This means that three over two π₯ plus nine π¦ minus 35 over two is less than zero.

We can rearrange this to get that nine π¦ is less than negative three over two π₯ plus 35 over two. Next, we can divide both sides of the inequality by nine and finally expand the brackets. And we find that the region for which the solution curves the differential equation are concave down is π¦ is less than negative one-sixth π₯ plus 35 over 18.

Moving on to part 3 of the question, we have that π¦ is equal to π of π₯ is a particular solution to the differential equation. Then, if π of one is equal to two, we have to find if π has a relative minimum, a relative maximum, or neither at π₯ is equal to one. Now, we know that a relative minimum or a relative maximum will occur when dπ¦ by dπ₯ is equal to zero. Since when dπ¦ by dπ₯ is equal to zero, this means that the slope of the curve is equal to zero too. Therefore, we have a critical point. And relative maximums and relative minimums occur at critical points.

Now, for the point at which weβre interested in, weβve been given the π₯-value, which is π₯ is equal to one. However, weβve also been told that π¦ is equal to π of π₯ is a particular solution which weβre interested in. Weβve also been told that π of one is equal to two. So we can say that when π₯ is equal to one, π¦ is equal to π of one which is also equal to two. So weβre trying to find the value of dπ¦ by dπ₯ when π₯ is equal to one and π¦ is equal to two.

We can simply substitute π₯ equals one and π¦ equals two into our equation for dπ¦ by dπ₯. When we rearrange this, we find that dπ¦ by dπ₯ is equal to one-half. We could have in fact used our answer to part 1 in order to find this value since we found dπ¦ by dπ₯ at this point for our slope field. So now, we have found the value of dπ¦ by dπ₯ and it is equal to one-half. Therefore, when π₯ is equal to one, for this particular solution, there is no critical point since dπ¦ by dπ₯ is not equal to zero. We can conclude that there is neither a local maximum or minimum at π₯ is equal to one.

Part 4, find the values of the constants π and π for which π¦ is equal to ππ₯ plus π is a solution to the differential equation.

In order to find the values of π and π, we can simply substitute our equation π¦ is equal to ππ₯ plus π into our differential equation. However, we first need to find dπ¦ by dπ₯. Since π¦ is equal to ππ₯ plus π, when we differentiate this, weβll get that dπ¦ by dπ₯ is equal to π. Now, we can substitute π¦ is equal to ππ₯ plus π and dπ¦ by dπ₯ is equal to π into our differential equation. We obtain that two π is equal to π₯ plus six multiplied by ππ₯ plus π minus 12.

Next, we can expand the brackets on the right-hand side. And then, since we have an extra π in the first two terms on the right-hand side, we can factor it out. Next, we simply need to choose π and π values such that this equation is true. We can start by comparing π₯ coefficients. On the left-hand side of the equation, we have no π₯ terms. Therefore, the coefficient of π₯ will be zero. And on the right-hand side, we have six π plus one π₯. Now, in order for this equation to be true, we need six π plus one to be equal to zero. Now, we can make π the subject of this equation. And so, weβve found the value of π is negative one-sixth.

Next, we can compare the constants on either side of the equation. On the left-hand side, we have two π. And on the right-hand side, we have six π minus 12. And again, in order for this equation to be true, we need these two things to be equal we have that two π is equal to six π minus 12. Now, we can substitute in the value of π which we just found, giving us that negative one-third is equal to six π minus 12. Rearranging, we find that six π is equal to 35 over three. And finally, we simply divide by six to obtain that π is equal to 35 over 18. Here, we have reached our solution. We found that π is equal to negative one-sixth and π is equal to 35 over 18. And so, we have that π¦ is equal to negative one-sixth π₯ plus 35 over 18.

Weβve now found answers to all four parts of the question. And these are as follows. For part 1, we have drawn the slope field as shown. For part 2, we found that d two π¦ by dπ₯ squared is equal to three over two π₯ plus nine π¦ minus 35 over two. And the region for which all the solution curves are concave down is π¦ is less than negative one-sixth π₯ plus 35 over 18. For part 3, we found that at π₯ is equal to one, this particular solution has neither a local maximum or local minimum. And for part 4, we found that the solution to the differential equation of the form π¦ is equal to ππ₯ plus π is π¦ is equal to negative one-sixth π₯ plus 35 over 18.