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# MATH 2320

Final review

1. Find the general solution of the dierential equation and use it to determine how solutions behave
as t → ∞.
y ′ + 7y = t + e−5t

:

1
1
t

+ e−5t + ce−7t
7 49 2
t
1
Solutions onverge to the fun tion y = − .
7 49
y=

2. Find the solution of the given initial value problem in expli it form
y ′ = (1 − 7x) y 2 ,

:
y=

7x2

y(0) = −

1
3

2
− 2x − 6

3. Find the solution of the given initial value problem in expli it form
sin 2x dx + cos 5y dy = 0,

:
y=

y 

π 
2

=

π
5 

1
π − arcsin 5 cos2 x
5

4. Determine without solving the problem an interval in whi h the solution of the given initial value
problem is ertain to exist. 

25 − t2 y ′ + 2ty = 8t2 ,

y(−8) = 1

:
(−∞, −5)

5. Determine without solving the problem an interval in whi h the solution of the given initial value
problem is ertain to exist. 

25 − t2 y ′ + 2ty = 3t2 ,

y(2) = −3

:
(−5, 5)

6. Classify the riti al (equilibrium) points as asymptoti ally stable or unstable.
dy
= y(y − 2)(y − 5),
dt

y0 ≥ 0

:(0, 0) is unstable, (2, 0) is asymptoti ally stable, and (5, 0) is unstable.
1

y). y4 = 4.9375 The exa t solution of the equation y + y ′ = 3 + 2t. and ( 7. dy = f (t. Answer : Given dy = f (t. y1 )h = y1 + m1 h and in general yn+1 = yn + f (tn . (0.5. by taking a small step on the tangent line orresponding to urrent value of the iteration. starting at y0 we build y1 using y1 = y0 + f (t0 . 0) is asymptoti ally stable. have to obtain the approximations at t1 = 0.5. y2 = y1 + m1 h = 1. 1. is given by y = 1 + 2t − e−t . Find the values of b for whi h the equation is exa t.5 = 2. Using Euler's method with h = 0. and then solve it using that value of b.75 Similarly we obtain m2 = 2. dy = y 2 (7 − y 2 ). Note that yn gives an approximation for the a tual solution y(t) evaluated at tn .5 and 2. is m0 = f (0.  8.5 = 1. y) = 3 + 2t − y and starting at t0 = 0. m1 is m1 = 3 + 1 − 1. 0) is semistable. dt y(t0 ) = y0 the Euler method onsists of building an approximation to the solution. unstable. 0 is unstable. y0 )h = y0 + m0 h The pro ess ontinues with y2 = y1 + f (t1 .5. 2 . 1. 9.5 nd approximate values of the initial value problem at t = 0.5.  ye7xy + x dx + bxe7xy dy = 0 Answer : 1 7xy x2 e + =c 7 2 b = 1.5 · 0. Using a step of size h in the input. t3 = 1. t4 = 2. 0) = 3. dt Answer √ −∞ < y0 < ∞ √ : − 7. t2 = t1 + h and tn = tn−1 + h = t0 + nh. t2 = 1. Classify the riti al (equilibrium) points as asymptoti ally stable.125. y3 = 3. m0 .5 + 2. using For the given dierential equation we have dt step h = 0.5 The next slope.5 = 2. Let y ′ = 3 + 2t − y and y(0) = 0.5. The rst slope.25.875 m3 = 2. and y1 = y0 + m0 h = 0 + 3 · 0. using an iterative method. yn )h = yn + mn h where t1 = t0 + h. or semistable.7.

Find the solution of the initial value problem 16y ′′ − y = 0. y(−4) = 1. 6) 16. Determine without solving the problem the longest interval in whi h the given initial value problem is ertain to have a unique twi e dierentiable solution. Use Euler's formula to express e3+9i into the form a + bi. Answer y(3) = 6. Consider the equation y ′′ − y ′ − 2y = 0 Note that y1 (t) = e−t and y2 (t) = e2t form a fundamental set of solutions. Answer : y= 5 −t/4 3e t/4 e − e 2e 2 13. y ′ (−4) = −1 . 1 + cos 2θ . y ′ (3) = −7 : (0. it is a solution. Answer : 0 14. Answer : e3 (cos 9 + i sin 9) = e3 cos 9 + e3 sin 9 3 . 17. xex Answer : x2 ex 15. Is y3 (t) = 3e−t + 2e2t also a solution of this equation? Answer : Yes. Solve the dierential equation Answer : 3x2 − 9y − y 3 dy = dx 9x + 3xy 2 9xy + xy 3 − x3 = c 12. Solve the dierential equation Answer x4 − 8y dy = dx x : x3 c + x8 12 y= 11.10. Find the Wronskian of the given pair of fun tions cos2 θ . Find the Wronskian of the given pair of fun tions x. t (t − 6) y ′′ + 2ty ′ + 5y = 5.

Answer : y = c1 e3t cos t + c2 e3t sin t 19. whi h from the given equation should equal ex sin x. The orresponding homogeneous equation has solution yh = (a1 + a2 x + a3 x2 + a4 x3 )ex sin e the hara teristi equation is (r − 1)4 = 0. Find the general solution of the given dierential equation y ′′ − 2y ′ − 3y = −3te−t . Find the general solution of the given dierential equation y ′′ − 6y ′ + 10y = 0. Find the solution of the given initial value problem 16y ′′ − 24y ′ + 9y = 0. 1 y = 3e 4 t + y ′ (0) = 5 11 3 t te 4 4 21. If we introdu e the dierential operator D = dx then the 4 x equation given is equivalent to (D − 1) y = e sin x and we are looking for a parti ular solution yp = u(x)ex . Answer : 1 5 y = c1 e− 4 t + c2 e 4 t 20. Then we see that (D − 1)4 yp = (D 4 u)ex = (c1 sin x + c2 cos x)ex . Answer.18. y ′ (0) = 1 11 3 1 1 2 cos 2t + sin 2t − + t2 + et 40 10 8 4 5 23. Answer : y = c1 e−t + c2 e3t + 3 −t 3 2 −t te + t e 16 8 22. Solve the given initial value problem y ′′ + 4y = t2 + 2et . so c1 = 1 and c2 = 0. yp = sin x ex 4 . Answer : y(0) = 3. We need to look for a parti ular solution of d the form yp = (c1 sin x + c2 cos x)ex . Sin e Dyp = (Du(x)) ex + u(x)ex we have (D − 1)yp = (Du)ex . Find a parti ular solution for the dierential equation y (4) − 4y (3) + 6y ′′ − 4y ′ + y = ex sin x. Answer : y=− y(0) = 0. Find the general solution of the given dierential equation 16y ′′ − 16y ′ − 5y = 0.

Answer 0<t< π 6 : y = c1 cos 7t + c2 sin 7t − 1 + sin (7t) ln |sec 7t + tan 7t|   2t. Answer : y = c1 et/2 + c2 tet/2 + 2t2 et/2 25. 4y ′′ + 49y = 49 sec2 7t. The initial onditions are Q(t0 ) = Q0 . pie ewise ontinuous or neither . Use the method of variation of parameters to nd the general solution of the dierential equation. 0 ≤ t < 1 27.24. nd the harge Q on the apa itor at any time t. Thus R E(t) L L Using I = Q dI + RI + = E(t) dt C dQ we obtain the dierential equation dt C LQ′′ + RQ′ + 1 Q = E(t) C for the harge Q. the voltage drop a ross apa itor is Q/C . If the initial harge on the apa itor is 0. Answer :  Q(t) = 0.15 × 10−6 oulomb and there is no initial urrent. 1 ≤ t < 2   0. 2 ≤ t ≤ 3 on the interval 0 ≤ t ≤ 3. Q′ (t0 ) = I(t0 ) = I0 . Answer : Continuous. Use the method of variation of parameters to nd the general solution of the dierential equation. A series ir uit has a apa itor of 2 × 10−6 farad and an indu tor of 0. and the voltage drop a ross indu tor is L dI/dt.15 × 10−6 cos 103 t Elementary laws of ele tri ity give that the voltage drop a ross resistor is IR. 4y ′′ − 4y ′ + y = 16et/2 . Sin e R = 0 and E(t) = 0 we have a homogeneous se ond order equation LQ′′ + Q =0 C 26. Determine whether f (t) = 4 − 2t.5 henry. 5 is ontinuous.

y ′ (0) = 14. Use the Lapla e transform method to solve the initial value problem y ′′ − 2y ′ + 2y = e−t . s>0 s3  f (t) = t2 − 8t + 32 u(t − 4). 0 ≤ t < 6 . π ≤ t y(0) = 5.28. y ′ (0) = 4  t. 0 ≤ t < π . y ′ (0) = 3. Use the Lapla e transform method to nd the solution of the initial value problem y ′′ − 4y ′ − 21y = 0. g(t) = 2 3. Answer : y(0) = 12. Find the Lapla e transform Y (s) = L{y} of the solution of the initial value problem ′′ ′ y + 16y = Answer ( 1. y ′ (0) = 1. f (t) = 2 t − 8t + 32. where Answer : y(0) = 6. 0. Find the solution of the given initial value problem y ′′ + y = g(t). 7 1 1 y = et sin t − et cos t + e−t 5 5 5 6 . 6 ≤ t   t t 1 t g(t) = + 3 − u(t − 6) = − (t − 6) u(t − 6). Find the Lapla e transform of the given fun tion ( 0. t<4 . : g(t) = 1 − uπ (t) Y (s) = 1 − e−πs 5s + 3 + 2 s + 16 s (s2 + 16) 30. Answer : t≥0 y(0) = 0. 31. y = 5e−7t + 7e3t 29. 2 2 2 2 y(t) = 1 1 (t − sin t) − u6 (t) [(t − 6) − sin(t − 6)] + 6 cos t + 4 sin t 2 2 32. 4 ≤ t Answer :  e−4s 2 + 16s2 F (s) = .

Answer   − x3 = 0  x1 3x1 + x2 + x3 = 1   −x1 + x2 + 2x3 = 2 : 37. Consider the ve tors 1 7 1 x1 = − . x3 = − 3 3 3 x(1) (t) =  2t2 4t  and x(2) (t) = linearly independent? Answer : On (−∞. Solve the system of equations or show that there is no solution.   0 2  or show that is singular.  4et et  . 8) ∪ (8. Compute the inverse of the matrix A = Answer  3i 3 2 −10i 36. Compute the inverse of the matrix . x2 = .33. In what intervals are x(1) and x(2)  e3t 38. ∞) . If A = Answer :  1 + i −6 + 3i 1 + 4i 2−i  and B =   : Answer  1 −4  9 4  :  −2 9 2 1  or show that is singular. nd AB −15 + 9i 33 + 63i −8 + i −7 − 8i 34. 1  −7 14 1 −2  0 1 1 7  2 15 1 7  1  0 −1 35. 0) ∪ (0. Find the general solution of the system x =    ′ Answer : x = c1 1 i 7 2 i −i 2 t e + c2   x 1 −i .

λ2 = 2.39. Des ribe the behavior of the solution as t → ∞. Find the eigenvalues and the orresponding eigenve tors of the matrix Answer :   1 v= 0  −1   −2 v= 1  0   0 v= 1  −1 λ1 = 1.  1 6   3 −4  e2t 41. 8 . Determine the Taylor series about the point x0 = 6 for the fun tion f (x) = x2 . λ3 = 3. Also determine 1−x the radius of onvergen e of the series. :   Answer 22 x= 5 1 1 x(0) = 7 e − 5 7t x1 → ∞ and x2 → ∞ as t → ∞. Find the general solution of the system   1 1 6 x′ =  1 6 1  x 6 1 1 Answer :      1 1 1 x = c1  1  e8t + c2  −2  e5t + c3  0  e−5t 1 1 −1  40. A= 1 −2 −4 −1 42. Answer : x2 = 36 + 12(x − 6) + (x − 6)2 1 43. Answer : ∞ X 1 (−1)n (x − 2)n = 1−x n=0 The radius of onvergen e is ρ = 1. Solve the initial value problem x′ =   8 −1 6 1 x. Determine the Taylor series about the point x0 = 2 for the fun tion f (x) = .   3 2 2 4 1 .

y ′ (0) = −7 7 8 y = 8 − 7x − 8x2 + x3 + x4 + .. 6 11 3 11 5 y2 (x) = x − x + x − .. Answer : y(0) = 8. . Find the re urren e relation for the oe ients of a power series of y about the given point x0 . regular 49.. 2! 3! 4!  + 10  x3 x4 + + . Find the rst ve nonzero terms in the solution of the given initial value problem. x = 5. . Find the rst three nonzero terms in ea h of the two linearly independent solutions of the given dierential equation (solve by means of power series about the given point x0 ). Answer x0 = 2 : 4 (n + 2) (n + 1) an+2 + 3 (n + 1) an+1 + (n + 5) an = 0 46.. 3! 4!  where a0 is an arbitrary onstant.  6 + x2 y ′′ − xy ′ + 12y = 0. 2 3 47. 4y ′′ + (x + 1) y ′ + 5y = 0. (x + 2)2 (x − 5) y ′′ + 3 (x − 5) y ′ − 2 (x + 2) y = 0 Answer : x = −2. Answer : x0 = 0 1 y1 (x) = 1 − x2 + x4 + . . y ′′ + xy ′ + 2y = 0. 36 288 45..44. Find all the singular points of the given equation and determine whether ea h one is regular or irregular. Show that the solution of the given dierential equation y ′ − y = 5x2 has the form y(x) = a0  x2 x3 x4 1+x+ + + + . . 48. irregular. x (x − 5) y ′′ + 2x2 y ′ + 8y = 0 Answer : x = 0 is a regular singular point 9 . Determine whether the point x = 0 is a regular singular point of the given dierential equation..

for every k ≥ 1. Bessel fun tions arise in the study of wave propagation (for example. One of the Bessel fun tions is ∞ J0 (x) = X (−1)k 22k (k!)2 k=0 x2k Find the radius and interval of onvergen e of the power series for J0 . ∞). 10 .50. 2k (k!)2 2 k=1 ∞ X (−1)k (2k) (2k − 1) x2k−2 . 2k (k!)2 2 k=1 2k − 1 xJ0′ (x) ∞ X (−1)k = (2k) x2k 2k (k!)2 2 k=1 x2 J0′′ (x) = ∞ X (−1)k (2k) (2k − 1) x2k 2k (k!)2 2 k=1 Sin e + = 1. waves on a ir ular drum). Dierentiate J0 twi e and show that y(x) = J0 satises the equation x2 y ′′ + xy ′ + x2 y = 0 Answer : Using ratio test we get R = ∞ and I = (−∞. the oe ient of x2k in the series orresponding to the left hand side of the 2k 2k equation be omes 0. By re indexing we have x2 J0 (x) = ∞ ∞ X (−1)k 2(k+1) X (−1)k−1 2k x = 2 2x 2(k−1) 2k 2 (k!) 2 ((k − 1)!) k=0 k=1 Taking derivative term by term gives J0′ (x) J0′′ (x) = 1 ∞ X (−1)k = (2k) x2k−1 .
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