Solutions to Math 53 First Exam April 20, 2010


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1 Solutions to Math 53 First Exam April 0, 00. (5 points) Match the direction fields below with their differential equations. Also indicate which two equations do not have matches. No justification is necessary. (The scale on each is x, y.) I II III IV V VI Equation I, II, III, IV, V VI, or none Equation I, II, III, IV, V VI, or none y = y II y = e y IV y = y + VI y = x + none y = y I y = x III y = x y V y = y none
2 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page of 9. (0 points) (a) Give an interval in which the initial value problem (4 t )y + ty = 3t, y( 3) = has a unique solution; give your reasoning. (Note: you are not being asked to solve the differential equation!) (0 points) There are two plausible solutions. In the first place we can write the ode as y + t 4 t y = 3t 4 t which is a linear first order ode. We can now cite Theorem.4. of the book, which guarantees the existence (and uniqueness) of a solution y = φ(t), defined on an open interval containing t 0 = 3 where the functions are continuous. But these rational functions of t are t 4 t and 3t 4 t continuous so long as t ±, so the open interval we seek is (, ). Alternatively, we can rewrite the ode in the form y = 3t 4 t ty 4 t With this formulation, one can write y = F (t, y), and then apply Theorem.4. of the book on existence & uniqueness of solutions to firstorder ivps. It suffices to check that F (t, y) and F y (t, y) are continuous in a neighbourhood of (t 0, y 0 ) = ( 3, ), which is easily done since F (t, y) and F y (t, y) are defined and continuous provided t ±. The result then states that there exists a solution y = φ(t) on some interval ( 3 h, 3 + h). This result, importantly, does not guarantee the existence of a solution y = φ(t) for all t (, ).
3 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 3 of 9 (b) Construct a firstorder linear differential equation whose solutions all approach the curve y = t as t. Then solve your ODE and show that the solutions have this property. (0 points) One way to construct an ode with the required property is to obtain first an ode all of whose solutions, Y = φ(t), have the property that lim t φ(t) = 0. One such is given by Y + Y = 0 which has general solution Y = ce t. We then write y c (t) = ce t + ( t ), and observe that y c (t) satisfies the ode y + y = ce t + ( t ) + ce t + ( t ) = t + t We have obtained a candidate ode: y + y = t t. There are several ways to solve this in full, using integrating factors or undetermined coefficients. We use the laziest method. Suppose φ(t) solves the ode; then since the ode is linear and the coefficient functions are continuous at all t, the domain of definition of φ can be assumed to be (, ). In particular, φ(0) = y 0 is defined. Consider the function ψ(t) = ( y 0 )e t + ( t ) We have ψ(0) = y 0 + = φ(0), and φ, ψ both satisfy the constructed linear first order ode. By the uniqueness part of the existence and uniqueness theorem for first order linear odes, they must coincide. All solutions of the ode therefore being of the form ce t + t, they all have the required property.
4 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 4 of 9 3. (5 points) A 000 gallon pool initially contains 500 gallons of water with 500 milligrams of a chemical mixed in. At noon, a solution containing 4 milligrams of chemical per gallon of water begins entering at a rate of gallons per minute, while the wellmixed liquid flows out of the pool at a rate of gallons per minute. Let C(t) denote the amount (in milligrams) of the chemical in the pool at t minutes after noon. (a) Write a differential equation satisfied by C(t), and give the value of C(0). (4 points) At time t, the rate at which chemical is entering the pool is 4 mg/gal gal/min = 8 mg/min, and the rate at which chemical is leaving the pool is gal/min C(t) mg 500 gal = C(t) 50 mg/min. Thus C satisfies the differential equation C = 8 C 50 (mg/min). The initial value is C(0) = 500 mg. (b) Solve for C(t), showing all steps; your answer should not depend on any unknown constants. (6 points) Separating variables, we get the equation dc 8 C/50 = dt, which is equivalent to dc 000 C = dt 50. Integrating on both sides, we get the equation which gives C 500 ds t 000 s = ds 0 50, ln 000 C 500 = t 50. Multiplying by ( ) and taking the exponential function on both sides, we get 000 C 500 = e t/50. It follows that 000 C 500 = ±e t/50, which gives C = 000 ± 500e t/50. Only the solution with the minus sign satisfies the initial condition C(0) = 500, so we have C(t) = e t/50.
5 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 5 of 9 (c) What happens to C(t) as t? Explain. ( points) When t, the term 500e t/50 0, and hence C(t) = e t/ Thus the amount of chemical in the pool approaches 000 mg as t. (d) Suppose that the situation is exactly as described above, except that the inflow rate is 3 gallons per minute (and the outflow rate remains at gallons per minute). Write a differential equation satisfied by C(t) in this situation. (Assume 0 t 500.) You do not need to solve this differential equation. (3 points) The amount of water in the pool at time t is 500 gal + t min (3 ) gal/min = (500 + t) gal. At time t, the rate at which chemical is entering the pool is 4 mg/gal 3 gal/min = mg/min. and the rate at which chemical is leaving the pool is Thus C satisfies the equation gal/min C(t) mg (500 + t) gal = C(t) t mg/min. C = C t (mg/min).
6 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 6 of 9 4. (0 points) Find a solution y(t) to the initial value problem yy + t = 0, y(0) = 3 and specify the solution s interval of definition (i.e. domain of definition). The equation is separable; we may rearrange both sides and integrate to find y dy = t dt so that y = C t for some C. (Alternatively, our original ODE can be rewritten as d dt (y + t ) = 0, so that as a function of t, y + t is a constant.) We can evaluate C by plugging in the initial condition y(0) = 3; we find C = 9/. Thus Solving for y, we obtain y = 9 t. y = ± 9 t The solution has two branches; we must pick the one with minus sign because y(0) = 3. Hence y = 9 t The interval of definition is ( 3, 3). Notice the endpoints ±3 do not belong to the interval, because y(t) is not differentiable at these points.
7 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 7 of 9 5. (0 points) Find a solution to the differential equation satisfying y() =. dy dx = yexy x xe xy We can rewrite the equation as Let M = ye xy + x and N = xe xy. Then (ye xy + x) + (xe xy ) dy dx = 0. M y = xye xy + e xy and N x = xye xy + e xy are equal, and so the equation is exact. We wish to find a function ψ(x, y) such that ψ x = M and ψ y = N. Integrating M with respect to x we find that ψ(x, y) = Mdx = exy + x + h(y). To find h(y) we use the condition ψ y = N: ψ y = xe xy + h (y) = N = xe xy to conclude that h (y) = 0, and so h(y) = c is a constant. Therefore ψ(x, y) = exy + x + c. Any solution y(x) satisfies ψ(x, y(x)) = c for some constant c. Therefore, or, setting c = (c c ), this is equivalent to exy + x + c = c e xy + x = c. The initial condition y() = is obeyed if e + = c. Therefore the solution can be written implicitly as e xy + x = e +. But this can be solved for y: e xy = e + x xy = ln(e + x ) y = x ln(e + x ).
8 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 8 of 9 6. (0 points) Consider the autonomous differential equation dy dt = y 9y/3. (a) One equilibrium solution is y = 0. Now determine all other equilibria, and for each nonzero equilibrium, classify it as stable or unstable; give reasoning. (You do not need to discuss the status of y = 0.) (4 points) Let f(y) = y 9y /3. We first notice that the autonomous ODE can be written dy dt = f(y) = y 9y/3 = y /3 (y /3 9) = y /3 (y /3 3)(y /3 + 3) and that the function f is continuous everywhere, but that f y = f (y) = 3y /3 fails to be continuous exactly on the line {y = 0}. As a consequence, away from this line the equilibria are exactly those values y so that f(y) = 0. By the above factorization, this gives the nonzero equilibria y = 7 and y = 7 (note that we get both signs). Each of these equilibria is unstable, since f is positive there (see also footnote, next page): f (7) = 3(7) /3 = 3 > 0, f ( 7) = 3( 7) /3 = 3 > 0. (b) Determine precisely the regions where the graph of a solution curve (y versus t) is concave up, and where it is concave down. Show your reasoning. (4 points) We use the chain rule to compute: d y dt = d dt f(y) = f (y) dy dt = f (y)f(y) = ( 3y /3 )(y 9y /3 ) = y /3 (y /3 3)(y /3 9). Thus, potential sign changes take place at y = ±3 3/ = ± 7 = ±3 3, and also at the three equilibria y = 0, ±7. These give six intervals of yvalues to check the product s sign; we find d y dt > 0 y = y(t) concave up for: 7 < y < 7, 0 < y < 7, y > 7; and d y dt < 0 y = y(t) concave down for: y < 7, 7 < y < 0, 7 < y < 7. (c) On the axes below, sketch several solution curves (y versus t) satisfying y(0) = y 0 for various values of y 0. You must sketch enough curves to clearly depict all of the different behaviors you determined in parts (a) and (b), and label your axis to include the key values you found in (a) and (b). (But: you do not need to find explicit solutions.) (4 points) In addition to the equilibria, at least four curves (one in each region formed by the three equilibria) should be drawn, and the location of inflection points (i.e., concavity change) clearly labeled. As an example, the curves depicted at right roughly correspond to initial values y 0 = 30, 5,,, 30, as well as the equilibrium values 0, ±7.
9 Math 53, Spring 00 Solutions to First Exam April 0, 00 Page 9 of 9 Footnote to (a): Recall why f (y 0 ) > 0 implies the equilibrium y = y 0 is unstable: this says that for y slightly less than y 0, then f(y) is negative, and so y is decreasing and thus moving down, i.e. away from the equilibrium. On the other hand, when y is slightly above the equilibrium y 0, then f(y) is positive, and so y is increasing and thus moving up, i.e. away from the equilibrium. (d) Find explicit expressions for three distinct functions y(t), each defined for all real values of t, and each satisfying the following initial value problem: dy dt = y 9y/3 y(3) = 0. for all t in (, ), (4 points) We note that y (t) = 0 is readily seen to satisfy both the ODE and the initial conditon and so is one possible solution. In order to get other solutions, we could separate variables and integrate (since any autonomous equation is separable). The integration, seemingly difficult, is greatly simplified via the substitution u = y /3. (Try this!) Alternatively, we could simply transform the original ODE using this change of variables u = y /3. We then have that u satisfies (by the chain rule): du dt = dy y /3 3 dt = 3 y /3 (y 9y /3 ) = (u 9). 3 It then follows that u(t) = 9 + Ce /3t. As a consequnce we have y(t) = ± (9 + Ce t/3) 3/ In order to have y(3) = 0, we must take C = 9e. But this gives a function defined only on t 3, since the halfpower fails to be defined when 9 + Ce t/3 < 0. However, it is not hard to check that the piecewise function: { (9 9e t/3 ) 3/ t < 3 y (t) = 0 t 3 is continuous and differentiable for all t and satisfies both the equation and the initial condition. The same can be said of the function y 3 (t) = y (t), which is clearly distinct from y (t). Thus y (t), y (t), y 3 (t) are three distinct solutions. (e) Discuss in precise terms how the phenomenon of part (d) can be reconciled with the existence and uniqueness theorem for nonlinear ODEs. (You can answer this question without having found a complete answer to part (d).) (4 points) As mentioned before, the function f(y) fails to have a continuous derivative at y = 0. This means that the existence and uniqueness theorem tells us nothing about the uniqueness of solutions that pass through points of the form (t, 0). In particular, there may be be multiple solutions to the ODE through a given point of (t, 0). Indeed, we give three such solutions through (3, 0).
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