### Video Transcript

Consider the differential equation five dπ¦ by dπ₯ equals π₯ π¦ minus three squared.

There are three parts to this question. Letβs start with part one and weβll look at parts two and three afterwards.

Part one) A slope field for the given differential equation is shown. Sketch the solution curve that passes through the point one, three and the one that passes through the point two, two.

Remember, slope fields help us to visualize solutions to differential equations, where each of these lines represents the value of the derivative at that point. Weβre asked to draw the solution curve that passes through the point one, three. If we find the point one, three, we can quickly see the direction in which the line proceeds. We end up with a horizontal line. And now, we want the solution curve that passes through the point two, two. We firstly find the point two, two. And we follow the direction of the lines to the left, and we follow the direction of the lines to the right.

So thatβs part one of our question. And as I said, we have two more parts to this question. So letβs go ahead and take a look at part two.

Part two) Let π¦ equal π of π₯ be the particular solution to the given differential equation with the boundary condition π of two equals two. Write an equation of the tangent to the graph of π of π₯ at π₯ equals two. Use your equation to approximate π of 1.7.

Remember, to find the equation of a tangent to a curve, we use the derivative as this gives us the slope of the tangent to the curve. So we want dπ¦ by dπ₯. We start with our differential equation. We can actually just rearrange our equation to get dπ¦ by dπ₯ by dividing each side of the equation by five. Remember, we want the gradient at the point where π₯ equals two. And we know that π¦ equals two at that point. So we evaluate dπ¦ by dπ₯ at two, two. This gives us that dπ¦ by dπ₯ equals two multiplied by two minus three squared all over five. This simplifies to two multiplied by negative one squared over five, which is just two over five. And remember, that is the slope of the tangent at the point two, two.

So if we use the general equation of a straight line, π¦ equals ππ₯ plus π. Remember here, that π is the gradient or slope and π is the π¦-intercept. We know that the slope π is two-fifths. And we know that the tangent line goes through two, two. So if we substitute π₯ equals two and π¦ equals two, we have that two equals two over five multiplied by two plus π. Two over five multiplied by two is four over five. So we can rearrange for π to find that π equals six over five. And so, the equation of the tangent at the point two, two is π¦ equals two-fifths π₯ plus six-fifths. So thatβs the equation of the tangent at the point two, two.

Weβre also asked to use our equation to approximate π of 1.7. So we use our equation of the tangent to approximate π of 1.7 by substituting π₯ equals 1.7 into the equation for our tangent. So if we worked this out, which you could do by converting all of the values to decimals, we approximate π of 1.7 to be 1.88.

So that was part two of our question. And finally, letβs have a look at part three.

Part three) Find the particular solution π¦ equals π of π₯ to the given differential equation with the boundary condition π of two equals two.

Weβll answer this by starting with our differential equation and separating the variables. We start by gathering the π¦s and dπ¦ on one side of the equation and the π₯s and dπ₯ on the other side. Multiplying both sides by dπ₯ and dividing by π¦ minus three squared gives us five over π¦ minus three squared dπ¦ equals π₯ dπ₯. And then, we integrate each side. Letβs firstly rewrite this bit as a negative exponent.

So on the left-hand side, we have the integral of five multiplied by π¦ minus three to the power of negative two. And weβre integrating that with respect to π¦. We could integrate this by recognizing that itβs the reverse of the chain rule. Or we could use substitution. Letβs go ahead with substitution and substitute π’ equals π¦ minus three. So we can replace π¦ minus three in our integral with π’. And so, we also need to replace the dπ¦. We can do that by differentiating π’ with respect to π¦. This gives us that dπ’ by dπ¦ equals one. Thatβs because π¦ differentiates to one and negative three β the constant β differentiates to zero. This rearranges to give us that dπ’ equals dπ¦. So we can just replace dπ¦ with dπ’ in our integral. So with the substitution of π’ equals π¦ minus three, we have the integral of five π’ to the power of negative two dπ’.

Remember the power rule for integrals: the integral of π₯ to the power π with respect to π₯ is π₯ to the power of π plus one over π plus one plus the constant of integration π. And so, the integral of five π’ to the power of negative two with respect to π’ is five π’ to the power of negative one over negative one plus a constant of integration. Thatβs weβll call π one. Dividing by negative one will just make the numerator negative. So we have negative five π’ to the negative one plus π one. And weβll put this back into fractional form: negative five over π’ plus π one. And remember, we substituted π’ equals π¦ minus three. So we need to put this back in.

So weβve integrated the left-hand side to be negative five over π¦ minus three plus the constant of integration. So now, letβs integrate the right-hand side. We need to integrate π₯ with respect to π₯. And we remember the power rule of integration, noting that π₯ on its own is the same as π₯ to the power of one. So we add one to the power and divide by the new power to get π₯ squared over two plus constant of integration, which weβll call π two. So our differential equation becomes negative five over π¦ minus three plus π one equals π₯ squared over two plus π two. We can combine these two constants and just call the constant π.

Remember that we were given the boundary condition π of two equals two. So when π₯ equals two, π¦ equals two. Substituting this condition gives us negative five over two minus three equals two squared over two plus π. Negative five over two minus three is negative five over negative one, which is just five. And two squared over two is four over two, which is just two. And we can solve for π to get π equals three. So we can substitute π equals three into our solution. And now, letβs solve for π¦.

We multiply through by π¦ minus three and then divide through by π₯ squared over two plus three. Letβs double the terms on the left-hand side to get rid of the fraction in the denominator. And we, finally, add three to both sides, which gives us the solution that π¦ equals three minus 10 over π₯ squared plus six.