Differential Equations Exam 1 Solutions

Exam 1 Solutions 1. (25 points) Short Answer: (a) State the general Existence and Uniqueness Theorem: Let y 0 = f (t, y), y(t0 ) = y0 . If f is continuous in an open rectangle containing (t0 , y0 ), then a solution (to the IVP) exists. If ∂f /∂y is continuous on an open rectangle containing (t0 , y0 ), then the solution is unique. (b) Given y 0 = y − y 2 , answer the following (WITHOUT solving the DE): i. If y(0) = 2, compute y 0 (0) and y 00 (0) We note that: y 0 (0) = y(0) − (y(0))2 = 2 − 4 = −2. For y 00 , we could either do: y 00 = y 0 − 2yy 0 at t = 2 ⇒ y 00 (0) = y 0 (0) − 2y(0)y 0 (0) = 6 Or, y 00 = ∂f 0 · y = (1 − 2y)y 0 at t = 2 ∂y ⇒ y 00 (0) = (1 − 2y(0))y 0 (0) = 6 ii. Determine where y(t) is increasing, decreasing, concave up, concave down. Use the phase plot, which is an upside down parabola intersecting the y axis at y = 0 and y = 1. From this, we see that: • • • • If If If If y < 0 or y > 1, y(t) is decreasing. 0 < y < 1, y(t) is increasing. y < 0, or 1/2 < y < 1, y(t) is concave down. 0 < y < 1/2, or y > 1, y is concave up. (c) For the following DEs, label them as: Linear, Separable or Exact (you need only choose one, but show your work). DO NOT SOLVE THEM. sin(x) − xy dy = i. dx x2 Rewrite as: 1 sin(x) dy + y= dx x x so this is linear. dy x + 3xy 2 ii. = dx 1 + x2 Rewrite as: x + 3xy 2 x dy = = · (1 + 3y 2 ) dx 1 + x2 1 + x2 so this is separable. dy x + ey iii. = dx xey Typo- the fraction should have been negative. In that case, the equation is exact. (Otherwise, it would have been none of the choices). 2. (10 points) Set up, but do not solve, the initial value problem that relates the rate of change of the amount of salt in the tank at time t. State the largest interval on which we can say that the solution is valid: “The capacity of the tank is 200 liters, and initially the tank contains 100 liters of pure water. Salt water solution, whose concentration is 2 kilograms per liter, is pouring into the tank at a rate of 6 liters per minute, and the well mixed solution is leaving at a rate of 5 liters per minute.” The input rate is constant: 2 · 6 = 12. The output rate: Let Q(t) be the amount of salt in the tank at time t. We will also need to find the amount of water in the tank at time t. 1 We see that the water level in the tank is rising at a rate of 1 L/min, so the water level is 100 + t gallons at t minutes. Put it all together for the output rate: 5 L Qlbs · min (100 + t)gallons We now have the IVP: dQ 5 = 12 − Q Q(0) = 0 dt 100 + t From the Existence and Uniqueness Theorem, t > −100. From the physical capacity of the tank, t < 100. Put these together, and our solution is valid on the interval: −100 < t < 100 3. (45 points) Solve. If there is no initial condition, give the general solution. You may leave your solution in implicit form. dy =0 dx This equation is exact. Once you found f (x, y), remember that the general solution is given by f (x, y) = c: 2y 2 + 3x2 y − xy 3 = C (a) (6xy − y 3 ) + (4y + 3x2 − 3xy 2 ) (b) xy 0 + 3y = 2x5 , y(2) = 1. This equation is linear. Be sure to rewrite in standard form first: y0 + 3 = 2x4 x The integrating factor is found: e3 R 1 x dx 3 = e3 ln(x) = eln(x ) = x3 Multiply both sides: x3 (y 0 + (3/x)y) = 2x4 · x3 Rewrite and integrate: (x3 y)0 = 2x7 ⇒ x3 y = 1 8 x +C 4 so that 1 5 C x + 3 4 x With the initial condition y(2) = 1, C = −56, so the final answer is: y= y= 1 5 56 x − 3 4 x (c) y 0 = xy 3 (1 + x2 )−1/2 This is separable: Z y −3 dy = Z √ x dx 1 + x2 For the right hand side of the equation, let u = 1 + x2 so that du = 2x dx. Therefore, Z p 1 1 −2 − y = u−1/2 du = u1/2 + C = 1 + x2 + C 2 2 Since we may leave the answer in implicit form, you could stop there. 2 4. (20 points) Create an autonomous differential equation so that y = ±2 are both stable equilibria, and y = 0 is unstable: (i) Create an approprate phase plot, (ii) Write the corresponding DE, (iii) give the general solution to the DE (Leave your answer in implicit form). If you had trouble with part (i), the phase plot, come by the office. For part (ii), we see that: y 0 = −y(y − 2)(y + 2) (You could leave an arbitrary constant in front, but that was not necessary). For part (iii), integrate using partial fractions, where: 1 −1 1 1 1 1 1 = · + · + · y(y − 2)(y + 2) 4 y 8 y−2 8 y+2 Now the solution will be: 1 1 1 − ln |y| + ln |y − 2| + |y + 2| = −t + C 4 8 8 You may leave it in this form. 3