4r 2 12r + 9 = 0. r = 24 ± y = e 3x. y = xe 3x. r 2 6r + 25 = 0. y(0) = c 1 = 3 y (0) = 3c 1 + 4c 2 = c 2 = 1

Transcription

1 Mathematics MATB44, Assignment 2 Solutions to Selected Problems Question. Solve 4y 2y + 9y = 0 Soln: The characteristic equation is The solutions are (repeated root) So the solutions are and Question 2 4r 2 2r + 9 = 0 r = 24 ± y = e 3x y = xe 3x = 3 y 6y + 25y = 0; y(0) = 3, y (0) = Soln: The characteristic equation is r 2 6r + 25 = 0 r = ( ) 6 ± = 3 ± 4i 2 Solutions are y = c e 3x cos(4x) + c 2 e 3x sin(4x) y (x) = c (3e 3x cos(4x) 4e 3x sin(3x)) + c 2 (3e 3x sin(4x) + 4e 3x cos(4x)) y(0) = c = 3 y (0) = 3c + 4c 2 = 9 + 4c 2 =

2 Soln c 2 = 2 y = 3e 3x cos(4x) 2e 3x sin(4x) Question 3. Solve y + k 2 y = 0. The characteristic equation is whose solutions are so the solution is r 2 + k 2 = 0 r = ±ki y(x) = C cos(kx) + C 2 sin(kx). Question 4. Find an ODE with solutions (a) 2e x, These functions are solutions of the equation y + ay + by = 0 e 5x if the characteristic equation is r 2 + ar + b = 0 with solutions r = and r = 5, in other wds (r )(r + 5) = 0 So the equation is r 2 + 4r 5 = 0. y + 4y 5y = 0. (b) e x cos(x), e x sin(x) These functions solve the ODE with characteristic equation whose roots are ± i, in other wds the characteristic equation is (r + + i)(r + i) = 0 Question 5. r 2 + 2r + 2 = 0. xy 2(x + )y + 4y = 0; We seek a solution of the fm y 2 = vy where (v ) + (p + 2(y ) /y )v = 0 2 y (x) = e 2x

3 So So v = (y ) 2 e p = e 4x e 2x e 2lnx = e 2x x 2 v = e 2x x 2 dx Integration by parts: e 2x x 2 dx = udw = uw wdu where dw = e 2x dx and u = x 2 so w = 2 e 2x and du = 2xdx. So e 2x x 2 dx = 2 x2 e 2x + xe 2x dx xe 2x dx = udw = uw wdu where dw = e 2x dx and u = x so w = 2 e 2x and du = dx so xe 2x dx = 2 xe 2x + e 2x dx = 2 2 xe 2x 4 e 2x. So x 2 e 2x dx = 2 x2 e 2x 2 xe 2x 4 e 2x. So the solution is y 2 = vy = ( 2 x2 2 x 4. In fact any constant multiple of this is also a solution: so y 2 = x 2 + x + 2 is a solution. Question 6. (x 2 + )y 2xy + 2y = 0; y (x) = x y + p(x)y + q(x)y = 0 3

4 where p(x) = The solution is y 2 (x) = v(x)y where Substitute w = x 2, = ln w + = ln(x 2 + ). So 2x x 2 +. v = y 2 e p(x)dx = x 2 e dw = 2xdx 2x x 2 + dx. 2x x 2 + dx = dw w + v = x 2 eln(x2 +) = x2 + x 2. Thus Question 7. v = + x 2, v = x /x. y 2 = vy = (x /x)x = x 2. x 2 y 2xy + (x 2 + 2)y = 0; y = x cos(x). We seek a solution of the fm y 2 = vy where v = y 2 e pdx f p(x) = 2x/x 2. Here e p(x)dx = e 2/xdx = e 2 ln x = x 2. So and v = x 2 cos(x) x2 = sec 2 (x), v = sec 2 (x)dx = tan(x). Thus y 2 = x tan(x) cos(x) = x sin(x). 4

5 Question 8. xy y + 4x 3 y = 0 y (x) = cos(x 2 ). The equation is y + p(x)y + q(x)y = 0 where p(x) = /x so pdx = ln x. y 2 = vy where v (x) = y 2 e pdx = x cos 2 (x 2 ). v(x) = = 2 x sec 2 (x 2 )dx du sec 2 (u) where u = x 2 = 2 tan(u) = 2 tan(x2 ). So the solution is y 2 = 2 tan(x2 ) cos(x 2 ) = 2 sin(x2 ). Question 9. x 2 y + xy 9y = 0; y = x 3 y + x y 9 x 2 y = 0 y 2 = vy where v satisfies v = y 2 e pdx = x 6 e ln(x) = x 7. So v = 6 x 6 and y 2 = 6 x 3. 5

6 Thus another solution is y = x 3. Question 0. where p(x) = 2x x 2 f x <. ( x 2 )y 2xy + 2y = 0 y + p(x)y + q(x)y = 0 p(x)dx = 2x x dx = 2 dw w where w = x 2. This integral equals ln w = ln( x 2 ). Hence y 2 = vy where v = y 2 e pdx = x 2 So x 2 v = Now we use partial fractions: It follows that A = B =, so dx x 2 ( x 2 ) = dx x 2 ( + x)( x) x 2 ( x 2 ) = A x + B 2 x = A Ax2 + Bx 2. 2 x 2 ( x 2 x 2 ( x 2 ) = x x 2 Now Hence C = D = /2 and x 2 = C x + D + x = C( + x) + D( x) ( x)( + x) ( x)( + x) = 2( x) + 2( + x). So x 2 ( x)( + x) = x 2 + x + + x. 6

7 So v = ln x + ln + x x = x + ln + x x. Question. (a) f(x) = x 3, g(x) = x 2 x so g(x) = x 3 if x > 0 but g(x) = x 3 if x < 0. W (f, g) = fg gf W (f, g) = 0 if x < 0 since f x < 0 f = g W (f, g) = 0 if x > 0 since f x > 0 f = g F x = 0, f = f = g = g = 0 so W (f, g) = 0 at x = 0. (b) If f and g were linearly dependent, there would be C, C 2 with f all x. But then C f(x) + C 2 g(x) = 0 C x 3 + C 2 x 2 x = 0 C + C 2 x/ x = 0 so C = C 2 and C = C 2. Hence C = C 2 = 0. (c) Theem (text, 8th edition) reads: Let y and y 2 be the solutions of L[y] = y + p(t)y + q(t)y = 0 where p and q are continuous on an open interval I. Then y and y 2 are linearly dependent on I if and only if W (y, y 2 )(t) is zero f all t in I. Alternatively, y and y 2 are linearly independent on I if and only if W (y, y 2 )(t) is never zero in I. Parts (a) and (b) do not contradict these Theems since the Theems are only valid under the hypothesis that f and g solve some differential equation L[f] = 0 and L[g] = 0. (Similar theems in the 9th edition are Theem and Theem These two theems would also yield a contradiction if f and g solved L[f] = 0 and L[g] = 0.) Question 2. A tpedo is travelling at a speed of 90 km/hour at the moment it runs out of fuel. If the water resists its motion with a fce proptional to the 7

8 speed and if km of travel reduces its speed to 60 km/hour, how far will it travel befe it stops? Soln: Let y(t) be the distance the tpedo has travelled. Let t = 0 be the time when it runs out of fuel. So y(0) = 0 dy (0) = 90km/h dt Let the fce of the water on the tpedo be b dy (proptional to the speed). dt So the equation of motion of the tpedo is m d2 y dt 2 + bdy dt = 0 d dt (dy dt ) + b dy m dt = 0 This is a first der equation f dy dy : Define v(t) =, then dt dt so dv dt + b m v = 0 dv v = b m dt ln v = b m t + C v = C 2 e b m t v(t) = C 3 e b m t. Introduce K = C 3, K 2 = b (two constants which are unknown). Then m y(t) = t 0 v(t) = K e K 2t t v(t )dt = K e K 2t dt = K K2t e t 0 = K ( e K2t ) K 2 K 2 km of travel reduces the speed to 60 km/h 8 0

9 So f T such that y(t ) =, dy dy (T ) = 60 and (0) = 90, we get K dt dt = 90 When K K 2 ( e K2T ) =, K e K 2T = 60 in other wds K K e K 2T = = 3 2 e K 2T = 3 2 K 2 T = ln( 3 2 ) But K ( e K2T ) = K ( 2 K 2 K 2 3 ) = K = 3 K 2 K = 3. K 2 The distance travelled by the tpedo befe stopping is lim y(t) = K = 3 t K 2 So the tpedo travels 3 km befe stopping. Question 3 Solve the difference equation y n+ = n + 2 n + 3 y n f the initial value y 0. Also describe the behaviour as n Soln: y = 2 3 y 0 y 2 = 3 4 y = 2 4 y 0 y 3 = 4 5 y 2 = 2 5 y 0 Conjecture: y n = 2 n + 2 y 0 9

10 Proof by induction: True f n = (base case) Suppose true f n. Then y n = 2 y n+2 0. But y n+ = n+2y n+3 n = 2 y n+3 0 so the statement is also true f n +. This completes the proof by induction. As n, y n 0. Question 4 A student deposits $500 in a bank account. The interest rate is 6 % compounded monthly. The student also deposits $0 in the account every month. How much money is in the account after 2 years? Solution: The balance in the n-th month is y n = ρy n + b where b = 0 (in dollars) and ρ =.005 (because 6 percent yearly = 0.5 percent compounded monthly). We also know that a 0 = $ 500 (initial deposit). So the solution is (as in textbook chapter 2.9 equation (4)) where ρ =.005 and b = 0. When n = 24 this gives (in dollars) y n = ρ n y 0 + ρn ρ b (200 0) = =