MAT 275 Test 1 SOLUTIONS, FORM A

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1 MAT 75 Test SOLUTIONS, FORM A The differential equation xy e x y + y 3 = x 7 is D neither linear nor homogeneous Solution: Linearity is ruinied by the y 3 term; homogeneity is ruined by the x 7 on the right-hand side The order of the differential equation 5 (y + 8y (sin xy y + y = ln(x + is B 3 Solution: There is a y in the equation but no y 3 For the differential equation dt = 3(y (y + (y + 5 3, the function y(t = 5 is C an unstable equilibrium solution Solution: If F (y = 3(y (y + (y + 5 3, then F (y < 0 when y is a little smaller than 5, and F (y > 0 when y is a little bigger than 5 4 [Assume that any indefinite integrals are solvable The differential equation y dx = ex y 4 sin x can be solved using D neither an integrating factor nor separation of variables Solution: In standard form, the differential equation is dx ex y = sin x y, and the y prevents the existence of an integrating factor In the form dx = ex y sin x y, it should be clear that the right-hand side cannot be factored into F (xg(y

2 5 The slope field of a differential equation y = F (x, y is shown below Determine the long-term behavior of the solution that satisfies y(0 = (The origin is indicated by a sign, and the points one unit away have been marked with a x y A lim x y(x = + 6 [5 points A tank contains ounces of salt and 00 gallons of water A salt water solution which contains 03 ounces of salt per gallon of water enters a tank at the rate of 6 gallons per minute The solution is mixed and drains from the tank at the rate of 7 gallons per minute Let S(t be the amount of salt (in ounces in the tank at time t Set up (but do not solve an initial value problem for S(t which models this situation Solution: The amount of water in the tank at time t is t 7t = 00 t, so ds dt = ( rate in ( rate out = 6 gal min 03 oz gal 7 gal min S oz 00 t gal = 8 7S 00 t Along with the information that there is originally ounces of salt, the following initial value problem is produced: ds dt = 8 7S 00 t S(0 = Grading for common mistakes: points for no initial condition; 5 points for assuming the volume is constant

3 7 [0 points Use separation of variables to solve the following differential equation dx = 4x3 e y 8 x e y + 5e y Solution: dx = 4x3 e y 8 x e y + 5e y = e y ( 4x 3 8 x + 5 e y = 4x3 8 x + 5 dx e y = 4x 3 8 x + 5 dx e y = x 4 8x ln 8 + 5x + C e y = x 4 + 8x ln 8 5x + C y = ln ( x 4 + 8x ln 8 5x + C y = ln ( x 4 + 8x ln 8 5x + C Grading: +4 points for each basic step Grading for common mistakes: 3 points for doing 8 x dx incorrectly 8 [0 points Use Euler s Method with a step size of to approximate y(3 for the initial value problem below = ((x + yπ/ 3 cos(xyπ dx y( = 4 Solution: The basic formula for Euler s Method is ( y a + = y(a + dx = y(a + [ ( (a + y(aπ 3 cos(ay(aπ (a,y The first iteration starts with a = and y(a = 4: y(5 = y( + [ ( ( + ( 4π 3 cos( ( 4 π = = 55 The next iteration uses a = 5 and y(a = 55: y(3 = y(5 + [ ( (5 55π 3 cos(5 ( 55 π = Grading: +4 points for the formula for Euler s method, +3 points for finding (the approximation to y(5, +3 points for finding y(3 Grading for common mistakes: points for calculating with degrees 3

4 9 [0 points Find an integrating factor and use it to solve the following initial value problem (3t + 4y + 3ty = t 3 + 0t y( = 4 9 Solution: This problem had a typo in it; the solution given below received full credit First, divide through by 3t + 4, so that the coefficient of y is : The integrating factor is y + 3t 3t + 4 y = t3 + 0t 3t = + 4 3t/(3t µ(t = e +4 dt = e ln(3t +4/ = 3t + 4 Now multiply both sides of the equation by the integrating factor: 3t + 4y 3t + 3t + 4 y = t3 + 0t 3t + 4 d [ 3t + 4 y = t3 + 0t dt 3t + 4 t 3 + 0t 3t + 4 y = 3t + 4 dt y = t 3 + 0t 3t + 4 3t + 4 dt Grading: +3 points for putting the equation into standard form, +3 points for finding the integrating factor, + points for multiplying by µ(t, + points for finding the general solution Grading for common mistakes: 3 points for not integrating correctly, 5 points for an incorrect integrating factor, 5 points for not dividing by 3t + 4 The original problem originally had 3t + 5 in front of the y, but this was changed so that the initial condition would make the square root turn out nicely The indefinite integral can be done, however: t 3 + 0t t 3 3t + 4 dt = + 6t 3t + 4 dt + 4t 4t(3t 3t + 4 dt = + 4 3t + 4 dt + 4t 3t + 4 dt = 4t (3t + 4 / dt + 4t (3t + 4 / dt = 4 9 (3t + 4 3/ (3t + 4 / + C, which makes y = 3t + 4 ( 4 9 (3t + 4 3/ (3t + 4 / + C = 4 9 (3t C 3t + 4, and after substituting the intial condition, C = 3 4

5 0 [0 points Find the largest interval I containing t = 0 such that the differential equation below is guaranteed to have a unique solution for all t in I (t + + ty = tan(πt dt Solution: First, divide the equation through by t + : dt + t t + y = tan(πt t + = sin(πt (t + cos(πt t tan(πt You need to find the largest interval containing 0 such that and are continuous at t + t + every number in that interval The places where these functions are discontinuous are t = (which is where t + = 0 and t = k + (where k is an integer (which is where cos(πt = 0, ie,, 3,,, 3,, so I cannot contain any of these points The biggest interval that works ( is, Graded on a point basis; ie, one mistake generally resulted in 7 points (total, two resulted in 5 points (total, etc 5

6 MAT 75 Test SOLUTIONS, FORM B The differential equation 5y + e x y y = 0 is C non-linear and homogeneous Solution: Linearity is ruined by the y term The order of the differential equation 4(y 3 + 5y(y + 8y = e x is A Solution: It contains a y but no y 3 For the differential equation dt = 3(y (y + (y + 5 3, the function y(t = is B a semi-stable equilibrium solution Solution: If F (y = 3(y (y + (y + 5 3, then F (y > 0 when y is a little smaller than, and F (y > 0 when y is a little bigger than 4 [Assume that any indefinite integrals are solvable The differential equation dx = x3 y + 4x 5 can be solved using B by finding an integrating factor but not by separating variables Solution: In standard form, the differential equation is dx x3 y = 4x 5, and is first order linear In the original form, it is difficult to see how to factor the right-hand side into the form F (xg(y

7 5 The slope field of a differential equation y = F (x, y is shown below Determine the long-term behavior of the solution that satisfies y( = (The origin is indicated by a sign, and the points one unit away have been marked with a x y C lim x y(x is a constant 6 [5 points A tank contains 5 ounces of salt and 500 gallons of water A salt water solution which contains 05 ounces of salt per gallon of water enters a tank at the rate of 8 gallons per minute The solution is mixed and drains from the tank at the rate of 5 gallons per minute Let S(t be the amount of salt (in ounces in the tank at time t Set up (but do not solve an initial value problem for S(t which models this situation Solution: The amount of water in the tank at time t is t 5t = t, so ds dt = ( rate in ( rate out = 8 gal min 05 oz gal 5 gal min S oz t gal = 4 5S t Along with the information that there is originally 5 ounces of salt, the following initial value problem is produced: ds dt = 4 5S t S(0 = 5 Grading for common mistakes: points for no initial condition; 5 points for assuming the volume is constant

8 7 [0 points Use separation of variables to solve the following differential equation dx = 5x y 5 x Solution: dx = 5x y 5 x = 5 x (y y = 5x dx y = 5 x dx ln(y = 5x ln 5 + C y = e 5x / ln 5+C = e 5x / ln 5 C y = + Ce 5x / ln 5 Grading: +4 points for each basic step Grading for common mistakes: 3 points for doing 5 x dx incorrectly 8 [0 points Use Euler s Method with a step size of to approximate y(4 for the initial value problem below = ((x + yπ/ 3 cos(xyπ dx y(3 = Solution: The basic formula for Euler s Method is ( y a + = y(a + dx = y(a + [ ( (a + y(aπ 3 cos(ay(aπ (a,y The first iteration starts with a = 3 and y(a = : y(35 = y(3 + [ ( (3 + ( π 3 cos(3 ( π = = 05 The next iteration uses a = 35 and y(a = 05: y(4 = y(35 + [ ( ( π 3 cos(35 05 π = Grading: +4 points for the formula for Euler s method, +3 points for finding (the approximation to y(35, +3 points for finding y(4 Grading for common mistakes: points for calculating with degrees 3

9 9 [0 points Find an integrating factor and use it to solve the following initial value problem (e t + ty + (e t + y = e t + t y(0 = Solution: First, divide through by e t + t, so that the coefficient of y is : The integrating factor is y + et + e t + t y = et + t e t + t = (e µ(t = e t +/(e t +t dt = e ln(et +t = e t + t Now multiply both sides of the equation by the integrating factor: (e t + ty + (e t + y = e t + t d [ (e t + ty = e t + t dt (e t + ty = e t + t dt = e t + t + C Since y(0 =, this means e t + t y = + C e t + t = C + 0 = C +, or C =, making y = e t + t + e t + t Grading: +3 points for putting the equation into standard form, +3 points for finding the integrating factor, + points for multiplying by µ(t, +0 points for finding the general solution, + points for using the initial condition Grading for common mistakes: 5 points for an incorrect integrating factor, points for saying the right-hand side was e t +, 3 points for not integrating correctly, 3 points for reading the original coefficient of y as e t +t (instead of e t +, 5 points for not dividing by e t + t 4

10 0 [0 points Find the largest interval I containing t = such that the differential equation below is guaranteed to have a unique solution for all t in I t dt t y = sec(πt Solution: First, divide the equation through by t: sec(πt + ( ty = = dt t t cos(πt You need to find the largest interval containing 0 such that t and sec(πt are continuous at every number in that interval The places where these functions are discontinuous are t = 0 and t t = k + (where k is an integer (which is where cos(πt = 0, ie,, 3,,, 3,, so I cannot contain any of these points The biggest interval that works is (, 3 Graded on a point basis; ie, one mistake generally resulted in 7 points (total, two resulted in 5 points (total, etc 5