Math 217 Fall 2000 Exam Suppose that y( x ) is a solution to the differential equation

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1 Math 17 Fall 000 Exam 1 Notational Remark: In this exam, the symbol x y( x ) means dy dx. 1. Suppose that y( x ) is a solution to the differential equation, x y( x ) F ( x, y( x )) y( x ) 0 y 0. Then y'(x 0 ) must equal: a) y 0 b) x 0 c) F ( x 0, y( x )) d) F ( x, y 0 ) e) F ( x, y( x ) ) f) F ( x 0, y 0 ) g) F x ( x 0, y 0 ) h) F y ( x 0, y 0 ) i) Can only be determined from the given information when the initial value problem is known to have a unique solution. j) Cannot be determined from the given information. Solution: f y'(x 0 ) F(x 0, y(x 0 )) F(x 0, y 0 ). For which F ( x, y ) is the direction field plot of the equation x y( x ) F ( x, y ( x )) in the window [-,]x[-,] the following plot:

2 a) y b) x c) x y d) x + y e) x + y f) x y g) xy h) 1 y i) 1 + x j) x 3 Solution: b This is easiest done by elimination. The slopes evidently depend x on but not on y. That eliminates all answers but (b) and (j). All the slopes are positive. That eliminates answer (j) which is negative for negative x. Only (b) remains. With a little examination it becomes clear that the slopes in view are consistent with this answer. 3. Suppose that y( x ) is the solution of the initial value problem x, x y( x) y( x ) y( 1) 3. Then y( ) is equal to: a) 4

3 b) 1 c) 8 d) 6 e) 4 f) g) 1 h) 0 i) - j) -6 Solution: b The equation is separable: x x y( x ) y( x ) 1 y Simplify the constant and solve using y( 1) 3: x y( x) y( x) 1 x dy d + 1 x C x ln( y ) ln( x) + C y( x ) e C x

4 y( x) C x 3 C y( x) 3 x y( ) 1 4. Find the general solution y( x ) of the differential equation x y( x ) 1 y( x ) + cos( x ). (The answer is in implicit y( x ) 3 form.) ln ( y( x) ) a) 4 sin( x ) b) y( x) C e ( 4 sin( x )) 0 c) y( x) C e ( 3 sin( x )) ln ( y( x) ) 0 d) 3 sin( x ) e) y( x) 3 y( x) 4 cos( x) + C f) y( x) C e ( 4 cos( x) ) 0 g) y( x) C e ( 3 cos( x )) 0 h) y( x) 3 y( x) 4 sin( x) + C i) y( x) 3 y( x) 4 C sin( x ) j) y( x) 3 y( x) 4 C cos( x) Solution: b x y( x) y( x ) + 1 cos( x) y( x) 3

5 In Bernoulli form with n -3: x y( x ) cos( x ) y( x ) cos( x ) y( x ) 3 Substitute w( x ) y( x ) ( 1 ( )) 3 : w( x ) 4 y( x ) 3 x x y( x ) Substitute the original equation: Substitute for y( x) w( x ) 4 ( ) x y x 3 y( x) + 1 cos( x) y( x ) 3 w( x ) 4 ( y( x) ) cos( x) x w( x ) 4 ( w( x ) + 1 ) cos( x) x This is a separable equation: Substitute w( x ) y( x) 4 1 d w + 1 w 4 cos( x) d x + C ln ( w + 1 ) 4 sin( x) + C e ( 4 sin( x) + C) 1 w( x ) y( x ) 4 e ( 4 ( ) ) sin x + C 1

6 This answer is equivalent to b. 5. Consider the differential equation + x y( x) y ( x ) e ( x3 ). x What is a sensible first step towards its solution? a) Give up. The Existence Theorem guarantees that there is no solution. y( x) b) Make the change of variable v( x) x. c) Make the change of variable v( x ) y( x ). d) Make the change of variable v( x ) y( x ) ( ). e) Multiply both sides of the equation by 0. f) Multiply both sides of the equation by x. g) Multiply both sides of the equation by x. h) Multiply both sides of the equation by x. i) Multiply both sides of the equation by e ( x 3 ). j) Substitute y( x) C e ( x3) and solve for C.

7 Solution: h dx x The equation is linear. A sensible first step is to multiply by the integrating factor e x. 6. Find an integrating factor for the equation x + x y( x ) y( x ) 3 x y( x ). 3 x) a) e( x b) x 3 3 x) c) x d) e( 3 x e ( ) e) e ( 3 x) f) 3 x e x g) 1 3 x Solution: h h) x i) 3 e ( 3 x ) x j) ln( x ) In linear form the equation is: + x y( x) 1 x 3 y( x) 0 1 d 3 x x Integrating_factor : e x ( e x ) 3 7. Find y( 1 ) if y( t ) is the solution of the initial value problem y( t) + y( t) e, t ( t ) y( 0 ).

8 a) 3 e b) 3 c) 3 e d) 3 e) e e f) e g) e h) 4 i) 4 e j) 4 e Solution: j ( dt) The given equation is linear with integrating factor e Use y( 0) e ( t ). y( t) + y( t ) e ( t) t t y( t) + y( t ) e ( ) t e ( ) to solve for the constant: t e ( t) e ( t) + y( t ) e ( t ) y( t) t t [ t ) e( y( t) ] e ( t ) y( t ) dt + C e ( t ) y( t ) t + C y( t ) y( t ) t + C C e ( t) t + e ( t) y( 1) 4 1 e

9 8. Let u( x) 1 y( x) and let w( x ) y( x ) in conjunction with the differential equation + x y( x) x y( x ) y( x ). Choose the correct statement. (Only one statement is correct. If you remember the theory, then you can answer without calculating.) a)the u substitution converts the differential equation to an equation of the type x u( x ) f( x ) b) The w substitution converts the differential equation to an equation of the type x w( x ) f( x ) c) The u substitution converts the differential equation to a separable equation. d) The w substitution converts the differential equation to a separable equation. e) The u substitution converts the differential equation to a linear equation. f) The w substitution converts the differential equation to a linear equation. g) The u substitution converts the differential equation to a homogeneous equation. h) The w substitution converts the differential equation to a homogeneous equation. i) The u substitution converts the differential equation into an identity. j) The w substitution converts the differential equation into an identity. Solution: e

10 The equation is Bernoulli with n. The appropriate strategy is to make the change of variable u( x ) y( x ) ( 1 n) or u( x ) 1. This substitution converts the equation to a linear equation, y( x ) (after which an integrating factor is summoned). y( x) 9. Let u( x) x y( x ) and let w( x) in conjunction with x the differential equation x y( x ) y( x) x + y( x ). Choose the correct statement. (Only one statement is correct. If you remember the theory, then you can answer without calculating.) a)the u substitution converts the differential equation to a Bernoulli equation. b) The w substitution converts the differential equation to a Bernoulli equation. c) The u substitution converts the differential equation to a separable equation. d) The w substitution converts the differential equation to a separable equation. e) The u substitution converts the differential equation to a linear equation. f) The w substitution converts the differential equation to a linear equation. g) The u substitution converts the differential equation to a homogeneous equation. h) The w substitution converts the differential equation to a homogeneous equation. i) The u substitution converts the differential equation into an identity. j) The w substitution converts the differential equation into an identity. Solution: d

11 The equation is homogeneous. The substitution w( x ) equation into a separable equation. y( x ) x is appropriate. It converts the Solve the differential equation x y( x ) y( x) x. a) y( x) x ( e x 1 + C ) e x + C b) y( x) ln( x ) 1 + C ln( x) c) y( x) e x 1 + C e x d) y( x) e x 1 + C e x + C e) y( x) e x + C e x + C f) y( x) x ( e x + C ) e x + C g) y( x) e x + C ( e x + C ) x h) y( x) ln( x ) + C ln( x) + C i) y( x) x ( ln( x) + C ) ln( x ) + C j) y( x ) ln( x) + C ( ln( x ) + C) x

12 11. Which of the following is the general solution of the differential equation y( x) e ( x y( x )) + x e ( x y( x )) x y( x) 0 a) y( x) C e x b) y( x ) C x + 17 c) y( x) C x e) y( x ) ln( C x ) f) y( x) e ( C x) d) The equation has no general solution. g) y( x ) sin( C x ) h) y( x ) cos( C x) sin x C i) y( x) C x j) y( x ) x ( e x + C )

13 1. Which of the following differential equations is not exact? a) x y 3 dx + 3 x y dy 0 b) cos ( x + y) dx + cos ( x + y) dy 0 c) y e ( x y ) dx + x e ( x y ) dy 0 d) x e ( x y ) dx + y e ( x y ) dy 0 e) ( x + 3 y) dx + ( 3 x + ) dy 0 f) ( 4 x y) dx + ( 6 y x) dy 0 g) ( x y + 3 x ) dx + ( x y + 4 y 3 ) dy 0 h) ( 3 x + y ) dx + ( 4 x y + 6 y ) dy 0 dy i) π dx + π 0 j) All of the above differential equations are exact.

14 dx 13. What is the general solution to dt 5 x ( x + )? a) x( t) C e t b) x( t) C e ( 10 t) 1 C e ( 10 t ) c) x( t ) cos( t) + C d) x( t) 5 C e ( 4 t) e ( 3 t ) e) x( t ) ( x( t ) + ) C e ( 5 t) f) x( t) 5 C sin( 10 t ) 1 C sin( 10 t ) g) x( t ) C + C e ( t) 1 h) x( t) i) x( t) arctan ( t + C ) π e ( 5 t) C + e ( 5 t ) j) The equation has no solutions.

15 14. Suppose a certain population satisfies the differential equation dp P ( P ). dt Suppose P( 0 ) Float ( 5, 0 ) What is the carrying capacity of this population? a) 5 b) 10 c) d) 100 e) f) g) 1776 h) 50000π i) j)

16 15. Suppose a certain population satisfies the initial value problem dp 50 P ( 6000 P ), P( 0 ) 000. dt Choose the statement that correctly describes the concavity of the solution curve. You do not have to know the explicit solution. The right side of the equation tells you where the derivative of P( t ) increases and decreases. a) The graph is concave up everywhere. b) The graph is concave down everywhere. c) There is an inflection point where P 000. d) There is an inflection point where P 500. e) There is an inflection point where P f) There is an inflection point where P g) There is an inflection point where P h) There is an inflection point where P i) There is an inflection point where P j) There is an inflection point where P Solution: e Since P ( 6000 P ) P P or P ( 6000 P ) 3000 ( P 3000 ),

17 > 16. Suppose that a nuclear powered rowboat is moving at 00 ft/sec when its onboard powerplant suddenly stops running, disconnecting power to the oars. Suppose that 10 seconds later it has slowed to 50 ft/sec. Assume the resistance it meets is proportional to its velocity. How many feet will the nuclear powered rowboat coast in total? a) 1000/ ln() b) 500/ ln(9) c) 600 / ln() d) 900 / ln(5) e) 000 / ln(8) f) 600 ln(1) g) 800 ln(17) h) 554 i) 700 j) It coasts around the world endlessly. Solution: a

18 We have F m dv m dv and F k v. Therefore k v. We can divide by the mass and call dt dt the new constant ρ. dv Thus, ρ v. The solution to this equation is v( t ) v( 0) e ( ρ t ). or v( t ) 00 e ( ρ t ). We dt also know that v( 10) 50. Therefore e ( 10 ρ ). Thus, e ( ρ ) v( t) t 10 t 10 1 as follows: e 4. The distance the boat coasts is d t ln and so t and dt and we work that out > Int(00*(1/4)^(t/10),t 0.. infinity) 00*Int(exp(t*ln(1/4)/10),t0..infinity); ( 1 / 10 t) dt 00 e ( 1 / 10 t ln( 4) ) dt ( 1 / 10 t) 0 dt ln( 4) 17. Suppose a magical huntsman fires a crossbow bolt directly up in the air at a speed of 30 ft/sec. Suppose gravity decelerates the bolt at a constant rate of 3 ft/ sec/sec. In how many seconds will the bolt return to the ground, striking the huntsman?

19 a) 15 b) 10 c) 5 d) 50 e) 0 f) 5 g) 37.5 h) 3 i) 64 j) Determine which of the pairs of functions in answers (a) through (i) is not linearly independent near any point x on the real line. If there is exactly one such pair, choose that as your answer. Otherwise, choose (j). a) f( x) x; g( x) 1 x b) f( x ) cos( x ) ; g( x ) sin( x) c) f( x) e x ; g( x ) d) f( x) x 3 ; g( x ) ln( x ) e) f( x ) tan( x ); g( x ) sec( x ) x

20 f) f( x ) x + 4 x + 5; g( x ) x + 3 x + g) f( x) e x ; g( x ) e ( x ) h) f( x ) cos( x ); g( x ) sin( x ) i) f( x) e x ; g( x ) e ( + ) j) More than one of the pairs above. Solution: i Let a 1 e and a -1. Then a 1 e x + a e ( x + ) is identically zero. So the pair of functions in (i) are linearly dependent and it may be verified that no other pair is. 19. Let y 1 ( x) x and y ( x ) tan( x ). The Wronskian W ( y 1, y ) is given by a) x tan( x) x sec( x) b) 1 c) x 3 cos( x) d) 1 + x + x + 3 x x e) x sec( x) x tan( x ) f) x tan( x ) sec( x )

21 g) x sin( x ) h) e ( x ) i) x tan( x) + x 3 ln ( sec( x) ) 3 j) x cot( x ) Solution: e The Wronskian is x x tan( x) x x tan ( x ) or x sec( x ) x tan( x ). 0. Suppose that the general solution to the equation y( x) 9 y( x ) is given by x y( x ) K 1 cos( 3 x) + K sin( 3 x ). Let y( x ) be a solution for which y( 0 ) 5 and D( y)( 0 ) 9. What is y π 9? a) b) + 5 c) d) e) 3 3 5

22 f) g) h) i) 0 j) Undefined Solution: b If y( x ) K 1 cos( 3 x) + K sin( 3 x ) then y( 0 ) K 1 cos( 0) + K sin( 0 ) or 5 K 1 + y( x ) 5 cos( 3 x) + K sin( 3 x ). Also D( y)( x ) 15 sin( 3 x ) + 3 K cos( 3 x ) so 9 15 sin( 0) + 3 K cos( 0 ). It follows that K 3. Therefore y π 5 cos π + 3 sin π or y π K 0. Thus,