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Consider the differential equation π¦ is equal to one over π₯ dπ¦ by dπ₯ minus three.
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Part i) Draw a slope field at the four indicated points in the given figure.
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Part ii) Let π¦ is equal to π of π₯ be a particular solution of the differential equation.
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Write an equation of the tangent to the graph of π at the point zero, two.
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Part iii) Find the approximate value of π of 0.02 using the equation of the tangent line in part two.
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Part iv) Find π¦ is equal to π of π₯ with π of zero is equal to two.
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For the first part of this question, weβre required to draw a slope field at the four points in the figure.
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These points are negative one, one; negative one, two; zero, two; and one, one.
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In order to draw the slope field for each of these points, we need to find the value of dπ¦ by dπ₯ at these points.
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Letβs start by writing our differential equation in terms of dπ¦ by dπ₯.
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We obtain that dπ¦ by dπ₯ is equal to π₯ multiplied by π¦ plus three.
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We start by substituting in π₯ equals negative one and π¦ equals one.
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We obtain that dπ¦ by dπ₯ is equal to negative four.
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We can write this value into a table.
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Next, we can use π₯ is equal to negative one, π¦ is equal to two.
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We obtain that dπ¦ by dπ₯ is equal to negative five.
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Next, we use π₯ is equal to zero, π¦ is equal to two, giving dπ¦ by dπ₯ is equal to zero.
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Finally, we use π₯ is equal to one, π¦ is equal to one, giving us the result that dπ¦ by dπ₯ is equal to four.
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Now, we have found the values of dπ¦ by dπ₯ at these four points.
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In order to draw the slope field at these four points, we simply draw a line with the gradient of dπ¦ by dπ₯ at the points.
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The slope field at negative one, one will have a gradient of negative four.
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At negative one, two, it will have a gradient of negative five.
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At zero, two, its gradient will be zero.
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And at one, one, its gradient will be four.
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So this completes the first part of the question.
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In part two, weβre told that π is a particular solution to the differential equation.
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And weβre required to find the equation of the tangent to the graph at the point zero, two.
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Looking at our slope field, we can see that at the point zero, two, dπ¦ by dπ₯ is equal to zero.
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Now, our tangent will be a straight line.
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So it will be of the form π¦ is equal to ππ₯ plus π, where π is the gradient of the tangent and π is the π¦-axis intercept of the tangent.
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We have just shown that at the point zero, two, the gradient dπ¦ by dπ₯ is equal to zero.
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Therefore, we know that π will be zero in our equation.
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So our tangent will be of the form π¦ is equal to π.
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Now, weβre finding the tangent of the graph at the point zero, two.
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Zero, two is on the π¦-axis.
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Therefore, zero, two will be the π¦-axis intercept of our tangent.
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At this point, the value of π¦ is two.
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Therefore, we can conclude that the equation of our tangent is π¦ is equal to two.
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In part three, we need to find the approximate value of π of 0.02 using the equation of the tangent line in part two.
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Now, our tangent in part two was tangential at the point zero, two.
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Here, the π₯-value is zero.
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In part three, weβre asked to approximate the value of π of 0.02.
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Here, the π₯-value is 0.02.
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Since 0.02 is very close to zero, this means that weβre able to use the tangent from part two in order to estimate the value of π of 0.02.
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In order to do this, we simply find the π¦-value on our tangent when π₯ is equal to 0.02.
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Now, the equation of our tangent is π¦ is equal to two.
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And itβs not depending on π₯.
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This means that π¦ hold the value of two regardless of the π₯-value.
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Therefore, we can use this tangent to approximate the value of π of 0.02 as two.
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For part four of the question, we need to find π¦ is equal to π of π₯ with π of nought equal to two.
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In this part of the question, we need to find a particular solution.
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Therefore, weβre going to need to solve our differential equation.
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Now, our differential equation is of the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by β of π¦, where in our case π of π₯ is equal to π₯ and β of π¦ is equal to π¦ plus three.
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What this means is we can use the separation of variables method in order to solve this differential equation.
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We can write our differential equation as dπ¦ over π¦ plus three is equal to π₯ dπ₯.
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And now, we can integrate.
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We have that the integral of one over π¦ plus three with respect to π¦ is equal to the integral of π₯ with respect to π₯.
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Weβre able to integrate π₯ with respect to π₯ quite easily.
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We simply increase the power of π₯ by one and divide by the new power.
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This gives us one-half π₯ squared.
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Next, we need to integrate one over π¦ plus three with respect to π¦.
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Now, we know that the integral of one over π¦ with respect to π¦ is equal to the natural logarithm of the absolute value of π¦.
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Now, we could use a π’ substitution here in order to show that our integral of one over π¦ plus three with respect to π¦ is equal to the natural logarithm of the absolute value of π¦ plus three.
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However, this is fairly straightforward.
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And so, we can just use this result here.
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We obtain that the natural logarithm of the absolute value of π¦ plus three is equal to one-half π₯ squared.
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Now, you may be wondering where the constants of integration for our indefinite integrals have gone.
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And thatβs because we havenβt added them in yet.
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We add a constant to each side of the equation: plus π on the left and plus π on the right.
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Now, we can move this constant both to the right-hand side.
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And we obtain a constant on the right of π minus π.
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However, since π minus π is also a constant, we can relabel this as π.
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Now that we have an equation for π¦ in terms of π₯, letβs make π¦ the subject of this equation.
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We can write both sides of the equation as powers of π.
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Now that the natural logarithm in the left-hand side of the equation is the inverse function of π, therefore these two functions cancel out with one another.
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Therefore, weβre left with the absolute value of π¦ plus three is equal to π to the power of a half π₯ squared plus π.
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Since we know that π is positive and it is simply being raised to a power, we know that the right-hand side of our equation is always positive.
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Therefore, we can lose the absolute value signs here, since the right-hand side of our equation is always going to be greater than zero.
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Next, we can split up the right-hand side of our equation.
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We write it as π to the power of one-half π₯ squared times π to the power of π.
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Now, π to the power of π is simply another constant since both π and π are constants.
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Therefore, we can rewrite this as another constant π.
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Now, we simply subtract three from both sides of the equation.
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And weβre left with π¦ is equal to ππ to the power of one-half π₯ squared.
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Now, we have nearly found the particular solution here.
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All that remains is to find the value of our constant π.
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Weβll be using information given in the question.
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And that is that π of nought is equal to two.
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π of nought is equal to two tells us that at π₯ equals zero, π¦ will be equal to two.
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Therefore, we substitute π₯ equal to zero and π¦ equal to two into our equation.
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We obtain that two is equal to π multiplied by π to the power of zero.
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π to the power of zero is simply one.
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Therefore, we find that π is equal to two.
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From here, we reach our solution which is that π¦ is equal two π to the power of one-half π₯ squared.
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Weβve now found solutions to all four parts of this question.
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Part one, we found the slope field for the four points in the figure.
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For part two, we found that π¦ is equal to two.
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For part three, we found that π of 0.02 is approximately equal to two.
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And for part four, we found that π¦ is equal to two π to the power of one-half π₯ squared.