Video: Solving a Separable Differential Equation, Sketching a Slope Field for It, and Finding the Equation of a Tangent to a Solution

Consider the differential equation 2π‘₯(d𝑦/dπ‘₯) βˆ’ 2(d𝑦/dπ‘₯) = 𝑦². i) In the given figure, sketch a slope field for the given differential equation at the indicated six points. ii) Let 𝑦 = 𝑓(π‘₯) be the particular solution to the given differential equation with the initial condition 𝑓(3) = 4. Write an equation of the tangent to the graph of 𝑓(π‘₯) at π‘₯ = 3. Use your equation to approximate 𝑓(3.1). iii) Find the particular solution 𝑦 = 𝑓(π‘₯) to the given differential equation with the initial condition 𝑓(3) = 4.

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Video Transcript

Consider the differential equation two π‘₯ timesed by d𝑦 by dπ‘₯ minus two times d𝑦 by dπ‘₯ equals 𝑦 squared. Part i) In the given figure, sketch a slope field for the given differential equation at the indicated six points.

Just to clarify, this question has two more parts, which we will answer after we have answered the first part. Let’s identify the coordinates of the six points indicated in the figure in part i. They are zero, two; zero, one; zero, zero; two, two; two, one; and two, zero. To sketch a slope field with the given differential equation at these six coordinates, we must draw short line segments at each of these six coordinators, where the slope of the line segment is equal to the value of d𝑦 by dπ‘₯ at the coordinate through which the line segment passes.

To calculate the value of d𝑦 by dπ‘₯ at each of the six coordinates, let’s rearrange the differential equation to make d𝑦 by dπ‘₯ the subject. Factoring d𝑦 by dπ‘₯, we obtain that two π‘₯ minus two multiplied by d𝑦 by dπ‘₯ equals 𝑦 squared. This means that d𝑦 by dπ‘₯ equals 𝑦 squared over two π‘₯ minus two. Substituting the coordinate zero, two into the expression for d𝑦 by dπ‘₯, we obtain two squared divided by two times zero minus two, which is equal to negative two.

Similarly, we find that the value of d𝑦 by dπ‘₯ at the coordinate zero, one is negative a half. The value of d𝑦 by dπ‘₯ at the coordinate zero, zero is zero. The value of d𝑦 by dπ‘₯ at the coordinate two, two is two. The value of d𝑦 by dπ‘₯ at the coordinate two, one is a half. And the value of d𝑦 by dπ‘₯ at the coordinate two, zero is zero.

Now that we have found the slopes of the line segments at each of the six coordinates, let’s draw the line segments on the figure provided. Let’s start with the coordinates zero, zero and two, zero. The value of d𝑦 by dπ‘₯ at each of the coordinates zero, zero and two, zero is zero. So the line segments at these coordinates are horizontal. The value of d𝑦 by dπ‘₯ at the coordinate two, one is a half.

Note that the line segment at this point would also pass through the coordinate zero, zero if it were extended, as we travel one unit upwards for every two units across to obtain a point on the line with slope equal to a half. The value of d𝑦 by dπ‘₯ at the coordinate zero, one is negative a half. This means that the line segment at this point is a mirror image of the line segment at the coordinate two, one.

Note that the line segment at zero, one would pass through the coordinate two, zero if extended. The value of d𝑦 by dπ‘₯ at the coordinate two, two is two. This means that, for every one unit across, we travel two units upwards. And so the line segment at this point would pass through the coordinate one, zero if extended.

The value of d𝑦 by dπ‘₯ at the coordinate zero, two is negative two. So the line segment at this point is the mirror image of the line segment at the coordinate two, two. Note that this line segment would also pass through the coordinate one, zero if extended.

This completes the sketch of the slope field for the given differential equation at the indicated six points. The short line segments in a slope field represent possibilities for tangent lines to solutions of the differential equation. This is useful as it allows us to see a visual representation of the curves that could result as solutions of the differential equation.

Let’s now move on to the second part of the question. Part ii) Let 𝑦 equals 𝑓 of π‘₯ be the particular solution to the given differential equation with the initial condition 𝑓 of three equals four. Write an equation of the tangent to the graph of 𝑓 of π‘₯ at π‘₯ equal three. Use your equation to approximate 𝑓 of 3.1.

In order to write an equation of the tangent to the graph of 𝑓 of π‘₯ at π‘₯ equal three, let’s recall one of the general forms of the equation of a line. The equation of a line can be written in the form 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š is the slope or gradient of the line and 𝑐 is the value of 𝑦 where the line crosses the 𝑦-axis. We will switch to using the letter 𝑑 here instead of 𝑦 so that we do not get mixed up with the variable 𝑦 given to us in the question.

In order to calculate the slope of the tangent to the graph of 𝑓 of π‘₯ at π‘₯ equal three, we need to calculate the gradient function of 𝑓, or in other words the first derivative of 𝑓 with respect to π‘₯ at π‘₯ equal three. We are told that 𝑦 equals 𝑓 of π‘₯ is a solution to the given differential equation. So the gradient function of 𝑓 is just d𝑦 by dπ‘₯ equals 𝑦 squared over two π‘₯ minus two.

We are also told that 𝑓 of three equals four. Substituting π‘₯ equals three and 𝑦 equals four into d𝑦 by dπ‘₯, we obtain the value four. So the slope of the tangent line in question is four. Since the point three, four lies on a tangent line, we can substitute π‘₯ equal three 𝑑 equals four into the equation 𝑑 equals four π‘₯ plus 𝑐 to find 𝑐. Doing so, we find that four equals four times three plus 𝑐, which implies that 𝑐 equals four minus 12, which is negative eight. So the equation of the tangent to the graph of 𝑓 of π‘₯ at π‘₯ equal three is 𝑑 equals four π‘₯ minus eight.

The next part of the question asks us to approximate the value of the function 𝑓 at π‘₯ equal 3.1 using this equation Substituting π‘₯ equals 3.1 into the equation 𝑑 equals four π‘₯ minus eight, we obtain four times 3.1 minus eight, which equals 12.4 minus eight, which equals 4.4. So the value of the function 𝑓 at π‘₯ equal 3.1 as approximated by the tangent to 𝑓 at π‘₯ equal three is 4.4.

Let’s now answer part iii of the question. Part iii) Find the particular solution 𝑦 equals 𝑓 of π‘₯ to the given differential equation with the initial condition 𝑓 of three equals four.

We previously rewrote the given differential equation as d𝑦 by dπ‘₯ equals 𝑦 squared over two π‘₯ minus two. We are now required to solve this differential equation for the particular solution 𝑦 equals 𝑓 of π‘₯ with the initial condition 𝑓 of three equals four. In order to do this, note that this is a separable differential equation. This means that we can separate all of the 𝑦 and d𝑦 terms to one side of the equation and all the π‘₯ and dπ‘₯ terms to the other side of the equation. Doing so, we obtain one over 𝑦 squared d𝑦 equals one over two π‘₯ minus two dπ‘₯. Separating the variables like so is the first step to solving a first-order separable differential equation.

The second step is to integrate both sides of the resulting equation. We can rewrite one over 𝑦 squared as 𝑦 to the power of minus two. To integrate 𝑦 to the power of negative two with respect to 𝑦, we just increase the exponent by one and divide by the new exponent. To integrate one over two π‘₯ minus two, we take the natural logarithm of the absolute value of two π‘₯ minus two and divide it by two, as per the formula.

Remember that the natural logarithm is not defined for negative arguments. So the absolute value sign around two π‘₯ minus two is necessary. Remembering the constants of integration, we obtain negative 𝑦 to the power of negative one plus 𝑐 one equals a half multiplied by the natural logarithm of the absolute value of two π‘₯ minus two plus 𝑐 two, where 𝑐 one and 𝑐 two are the constants of integration.

Collecting the constants of integration together on the right-hand side then multiplying both sides of the equation by 𝑦 and then dividing both sides of the equation by a half multiplied by the natural logarithm of the absolute value of two π‘₯ minus two plus 𝑐 two minus 𝑐 one. We obtain that 𝑦 equals negative one over a half multiplied by the natural logarithm of the absolute value of two π‘₯ minus two plus 𝑐 two minus 𝑐 one.

To simplify the expression on the right-hand side, we multiply the numerator and denominator by two. We can denote the constant two times 𝑐 two minus 𝑐 one with the letter 𝑐. This completes the third step to solving a first-order separable differential equation, which is to write the general solution in the form 𝑦 equals 𝑓 of π‘₯.

The solution we have found is called a general solution because it contains the constant 𝑐, which is undetermined. The fourth and final step to solving a first-order separable differential equation is to use the initial conditions, if given, to find the particular solution. A particular solution should not contain any undetermined constants.

We will use the initial condition π‘₯ equal three, 𝑦 equals four, which is given to us in the question, to determine the value of 𝑐 and therefore find a particular solution to the differential equation given to us in the question. Substituting π‘₯ equal three and 𝑦 equals four into the general solution, we find that four equals negative two over the natural logarithm of four plus 𝑐.

Multiplying both sides of the equation by the natural logarithm of four plus 𝑐 then subtracting four times the natural logarithm of four from both sides and then dividing both sides by four. We obtain that 𝑐 is equal to negative a half minus the natural logarithm of four. Substituting this value of 𝑐 into the general solution 𝑦, we obtain negative two over the natural logarithm of the absolute value of two π‘₯ minus two minus a half minus the natural logarithm of four. This is the particular solution 𝑦 equals 𝑓 of π‘₯ to the given differential equation with the initial condition 𝑓 of three equals four.

Let’s rewrite this particular solution in a nicer way. Using the fact that the difference of logarithms is equal to the logarithm of the quotient, we can rewrite the natural logarithm of the absolute value of two π‘₯ minus two minus the natural logarithm of four as the natural logarithm of the absolute value of two π‘₯ minus two divided by four.

Next, we multiply the numerator and denominator of the bigger quotient by two and simplify the argument inside the natural logarithm. Now using the logarithm power rule, we can move two from the front of the natural logarithm to the exponent of the argument of the logarithm. Since squaring an absolute value makes the absolute value sign redundant, we can get rid of the absolute value sign from the argument of the natural logarithm. Taking the negative sign, which is outside the whole quotient, inside the denominator, we can rewrite their particular solution as four over one minus the natural logarithm of π‘₯ minus one all squared divided by four. This completes the answer to the third and final part of the question.

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