Differential equations #3 - Questions and Answers

Question 1 verify that each given function is a solution of the differential equation ty' - y = t²; y = 3t + t² Answer 1 Step 1 We have differentiated y(t) dy/dt = (3t + t²) = 3 + 2t Step 2 Plug the values of y(t) and dy/dt in the given differential equation t(3 + 2t) - (3t+t²) = t² 3t + 2t² - 3t - t² = t² y(t) has satisfied the equation Thus y(t) is a solution of given differential equation. Result y is a solution of given differential equation Answer 2 Step 1: First, let's find the derivative of y with respect to t. Using the power rule, we have: dy/dt = d/dt (3t+t²) = 3+ 2t Step 2: Next, we substitute y and dy/dt into the differential equation: t*dy/dt - y = t(3 + 2t) - (3t + t²) = 3t + 2t² - 3t - t² = t² We observe that the expression simplifies to t², which is indeed equal to the right-hand side of the differential equation. Since the left-hand side of the differential equation, ty' - y, equals the right-hand side, t², we can conclude that the function y = 3t + t² is indeed a solution of the differential equation ty' - y = t². Answer 3 Let's start by substituting y = 3t + t² into the differential equation: t * dy/dt - y = t² Now, let's calculate the derivative of y with respect to t. The derivative of 3t is 3, and the derivative of t² is 2t. Applying the power rule of differentiation, we have: dy/dt = d/dt (3t + t²) = 3 + 2t Substituting this derivative back into the equation, we get: t(3 + 2t) - (3t + t²) = t² Expanding the terms, we have: 3t + 2t² - 3t - t² = t² Simplifying the equation, we find: t² = t² Since the left side of the equation is equal to the right side, we can conclude that y = 3t + t² is indeed a solution to the given differential equation ty' - y = t². Therefore, the function y = 3t + t² satisfies the differential equation. Answer 4 To verify that each given function is a solution of the differential equation ty' - y = t², we need to substitute the given function y into the equation and see if it satisfies the equation for all values of t. Let's start by substituting y = 3t + t² into the differential equation: t * (dy/dt) - (3t+t²) = t² Now, we need to differentiate y with respect to t to find dy/dt: y = 3t + t² dy/dt = 3 + 2t Substituting this back into the equation: t(3 + 2t) - (3t+ t²) = t² Simplifying this equation gives us: 3t + 2t² - 3t - t² = t² t² = t² As we can see, both sides of the equation are equal, so the given function y = 3t + t² is indeed a solution of the differential equation ty' - y = t². Therefore, we have verified that y = 3t + t² is a solution of the differential equation. Question 2 Population y grows according to the equation dy/dt = ky, where is k - constant and t is measured in years. Of the population doubles every ten years, then the value of k is ? Answer The population growth can be measured as: x(1+r) So if it doubles every ten years: 2x = x(1+r)¹⁰ 2 = (1+r)¹⁰ ln 2 = 10ln (1+r) (1/10)ln 2 = ln (1+r) ln (1+r) = k k = (1/10)ln 2 ≈ 0.0693 Question 3 The problem question is: dy/dt = ay + by² we have to sketch the graph f(y) versus y, determine the critical points, and classify each one as asymptotically stable or unstable. Thing is, how do you get the critical points? Answer dy/dt = f(y) = ay + by² we take a = 2 and b = 4 and draw the graph of f(y) versus y [Image of a parabola going through the origin opening downwards] for the critical points we put f(y) = 0 ay + by² = 0 y(a + by) = 0 y= 0 and y = -a/b (critical points) Now we find second derivative that is d²y/dt² = (d/dy) (ay + by²) * dy/dt = (a + 2by) (ay + by²) Second derivative will be zero when y = -a/(2b). y = 0, y = -a/b But y = 0 and y = -a/b are the solution of the equation so we take y = -a/(2b) (inflection point) Now from the graph drawn above, we have dy/dt > 0 for y < -a/b, so y is increasing dy/dt < 0 for -a/b > 0 for y > 0 so y is decreasing and concavity changes at y = -a/(2b) The graphs of solutions of the equation must have the general shape as follows [Image of three different graphs, all starting from the left and approaching two parallel horizontal lines] Since the solutions diverge from the line y = 0 and converge to the line y = -a/b so y = 0 is an unstable point and y = -a/b is stable Question 4 Suppose that a population develops according to the logistic equation dP/dt = 0.05P - 0.0005P² where t is measured in weeks. What is the carrying capacity? What is the value of k? Answer Logistic differential equation: dP/dt = kP(1 - P/M) where M is carrying capacity Hence, to obtain M and k we need to rewrite the given equation: dP/dt = 0.05P - 0.0005P² = 0.05P(1 - 0.01P) = 0.05P(1 - P/100) Therefore we have: dP/dt = 0.05P(1 - P/100) And from this equation we get: k=0.05 M = 100 Question 5 Explain what is the difference between implicit and explicit solutions for differential equation initial value problems. Answer 1 You are attempting to solve for the function y=f(x) when resolving a differential equation (x). An equation of the kind would be a clear answer. If you discover an equation with only two variables, x and y, but you are unsure of how to solve for y, then you have found an implicit answer. After doing your study, you could discover that y=sin(y)+x or a similar equation. However, there isn't an instantly apparent method to reduce this to y=f(x) form. Once there are no more derivatives, this does uniquely define a solution to the IVP. Answer 2 Step 1: An initial value problem (IVP) for a differential equation involves finding a solution that satisfies both the differential equation and a set of initial conditions. There are two types of solutions for IVPs: implicit solutions and explicit solutions. Implicit Solution: An implicit solution is represented by an equation in which the dependent variable and its derivatives are related in a non-explicit form. In other words, the dependent variable is not expressed explicitly as a function of the independent variable. Instead, the equation may involve the dependent variable and its derivatives on both sides, without isolating the dependent variable on one side. For example, consider the first-order ordinary differential equation: dy/dx = f(x, y) An implicit solution to this equation would be an equation of the form: F(x, y) = C where F is a function involving x, y, and their derivatives, and C is a constant determined by the initial conditions. The implicit solution does not explicitly provide the functional relationship between x and y. Explicit Solution: Step 2: In contrast, an explicit solution expresses the dependent variable explicitly as a function of the independent variable. It provides a direct relationship between x and y, without involving derivatives of y on the left-hand side of the equation. Using the same example of a first-order ordinary differential equation: dy/dx = f(x, y) An explicit solution would be of the form: y = g(x) where g is a function that directly expresses y in terms of x. This type of solution allows for a clear and straightforward understanding of the relationship between the independent and dependent variables. Answer 3 An implicit solution for a differential equation is expressed in terms of an equation that relates the dependent variable y and the independent variable x, without explicitly solving for y as a function of x. In other words, the relationship between x and y is not explicitly given. Instead, the equation typically involves both x and y, and it can be difficult or even impossible to isolate y on one side of the equation. Implicit solutions are commonly represented as equations in the form F(x, y) = 0, where F is a function that incorporates both x and y. On the other hand, an explicit solution for a differential equation explicitly expresses y as a function of x. It provides a clear relationship between x and y, allowing us to determine the value of y for any given x. Explicit solutions are typically represented in the form y = f(x), where f is a function that directly relates x to y. To summarize, the main distinction between implicit and explicit solutions lies in the way the relationship between x and y is presented. Implicit solutions involve an equation that relates x and y without explicitly isolating y, while explicit solutions express y as a function of x.