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Unformatted text preview: Chapter 1. Introduction 1. The differential equation is second order, since the highest derivative in the
equation is of order two. The equation is linear, since the left hand side is a linear
function of y and its derivatives. 3. The differential equation is fourth order, since the highest derivative of the
function y is of order four. The equation is also linear, since the terms containing
the dependent variable is linear in y and its derivatives. 4. The differential equation is ﬁrst order, since the only derivative is of order one.
The dependent variable is squared, hence the equation is nonlinear. 5. The differential equation is second order. Furthermore, the equation is nonlinear,
since the dependent variable y is an argument of the sine function, which is not a
linear function. 7. y1(t) = 61" => = y1”(t) 2 et. Hence yl” — y1 = 0. Also, y2(t) z cosh 75 => = sinh t and yz”(t : cosh 15. Thus y2” — y2 = O. 9. y(f) 2 3t + t2 :> y’(t) = 3 + 2t. Substituting into the differential equation, we
have 15(3 + 275) — (375 + 152‘) = .375 + 2t2 — 3t — t2 = t2. Hence the given function is a
solution. 10. 3/105) = t/3 => gm) = 1/3 and Eli/(t) = was) = mt) = 0. Clearly. at)
is a solution. Likewise, y2(t) = e_t + 15/3 : = —e_t + 1/3, yé/(t) = e_t ,
yé”(t) = —e_t, yé’”(t) 2 6‘4. Substituting into the left hand side of the equation,
we ﬁnd that e“ + 4(—e_t) + 3(e‘t + t/3) = e_t — Ale—t + 36” +t = t. Hence both
functions are solutions of the differential equation. 11. y1(t) = t1/2 z> = t‘1/2/2 and yl”(t) = —t”3/2/4. Substituting into the
left hand side of the equation, we have 2t2(_t—3/2/4) + 3t(t~1/2/2) _ tl/Z : _t1/2/2 + 3t1/2/2 _ t1/2
= 0 Likewise, yg (t) = 73—1 => 3/2/05) 2 ——t_2 and yé’Oﬁ) = 2t_3. Substituting into the left
hand side of the differential equation, we have 2t2(2t_3)+3t(—t 2)—t 1:47: 1—3t 1—t 1:0. ’ Hence both functions are solutions of the differential equation. 12. y1(t) = 75—2 :> = ~—2t“3 and yl”(t) = 6 r4. Substituting into the left hand
side of the differential equation, we have l0
[0 t2(6t—4)+5t(—2t 3)+4t :67: 2—107: 2+4t =0. mer 2memmwwmrm mmmmmmawwmx “WmmmmwwmtmwWWWmmmmw.WWMWWWMWWWWWMmmmmmmmmewmmwwmm 2.1 27 16. The integrating factor is Mt) 2 6f i‘” = t‘2 . Multiplying both sides by Mt), the
equation can be written as (t2 y)’ = cost. Integrating both sides of the equation
results in the general solution y(t) = sin t/t2 + ct“? Substituting t = 7r and setting
the value equal to zero gives 0 = 0. Hence the speciﬁc solution is = sin t / t2.
17. The integrating factor is Mt) = e‘gl', and the differential equation can be
written as (e—2t y)’ = 1. Integrating, we obtain 8—2t y(t) = t + c. Invoking the
specified initial condition results in the solution y(t) = (t + 2)€2t. 19. After writing the equation in standard form7 we ﬁnd that the integrating factor
is Mt) = ef %dt = t4 . Multiplying both sides by Mt), the equation can be written as
(t4 y)’ = te‘t . Integrating both sides results in fig/(t) : —(t + 1)::—t + c. Letting
t = —1 and setting the value equal to zero gives (2 = 0. Hence the specific solution
of the initial value problem is y(t) = —(t'3 + t_4)e_l’. 21.(a) ‘4 é? ﬂ???
:i/’ //ff
;4 . x4/
‘0 f _’ lie 1"
n2 xx“ , x2!
‘/z—\\\.\ xi/f
/~*“*~\\\\‘—v~“/f
Hr—T‘u\\4\\\u{—"/:’
afﬁx“ \W’ '
¥%&\\xx\m»/ / _
Jxxxxkxhwex /,%~\
ﬁ‘c‘cttttitii‘ﬂ 5:“:
2 ;m\\txx\\\mw xxax\
. wk\\kMM\\kﬂwxwwx\
hittitﬁt‘crﬁt‘ql‘
a \Rﬁkkkilkxw x\\ The solutions appear to diverge from an apparent oscillatory solution. From the
direction field, the critical value of the initial condition seems to be a0 = —1 . For
a > —1, the solutions increase without bound. For a < —1, solutions decrease without bound. (b) The integrating factor is Mt) = €”t/2. The general solution of the differential
equation is y(t) = (8 sint — 4 cos t) /5 + act/2. The solution is sinusoidal as long as
c = 0. The initial value of this sinusoidal solution is a0 = (8sin(0) — 4cos(0))/5 = —4/5. ((3) See part  32 Chapter 2. First Order Differential Equations If this satisfies the differential equation then
1/ +p(t):l/ = A’(t)6‘f “W = 905)
and the required condition follows. (G) Let us denote Mt) = efpmdt. Then clearly A(t) = fp,(t)g(t)dt, and after sub—
stitution y = f p,(t)g(t)dt  (1/,u(t)), which is just Eq. (33). 40. We assume a solution of the form
y = A(t)e_f1?dt = Ame—1‘” = Amt—1,
where A(t) satisﬁes A’ (t) = 375 cos 215. This implies that __ 3cos2t 3tsi112t
“ 4 + 2 A(t) and the solution is .
3 cos 2t 3 S111 275 + +E
4t 2 t' y: 41. First rewrite the differential equation as ,+2 sint
i —z=
J tJ t Assume a solutionof the form
y = A(t)e‘f W = Amt2,
where A(t) satisﬁes the ODE
A’(t) = t sin t.
It follows that A(t) = sin 13 ~ 75 cos t + c and thus y = (sin 75 t cos t + c)/t2 . 5"
N l 2. For 1' 75 —1 , the differential equation may be written as
ydy = [932/(1 + elm. Integrating both sides, with respect to the appropriate variables, we obtain the relation y2/2 =%lnl1 +1'3l +c. That is, = ::‘/%111l1+1'3 +0. 3. The differential equation may be written as y‘gdy = ~sin (balm. Integrating
both sides of the equation, with respect to the appropriate variables, we obtain the
relation —y_1 = cos 1' + c. That is, (c — cos = 1, in which 0 is an arbitrary
constant. Solving for the dependent variable, explicitly, y(:r) : 1/(c — cos . 5. Write the differential equation as cos‘2 2y dy = cos2 :L' dZL', or sec2 2y dy = cos2 1' dm.
Integrating both sides of the equation, with respect to the appropriate variables,
we obtain the relation tan 2y 2 sin .t' cos :3 + :L‘ + c. i
y
‘3‘ 2.6 65 2.6 1. ]\/[(.r,y) = 22: + 3 and N($,y) = 9y — 2. Since JVIy = NH; 2 0, the equation is
exact. Integrating JVI with respect to :8, while holding y constant, yields ¢(5L‘, y) =
m2 + 3m + My). Now wy = h’(y), and equating with N results in the possible
function My) 2 y2 — 2y. Hence 1/1(a:, y) = 1'2 + 3.7: l y2 — 2y, and the solution is
deﬁned implicitly as 332 + 39; + y2  2y = c. 2. ]\/I(:c, y) = 9a: + 4y and N(.r, y) = 2:3  2y. Note that jl/Iy 7E NaU , and hence the
differential equation is not exact. 4. First divide both sides by (21y + 2). We now have ll/I(m, y) = y and N(a;, y) = x.
Since ll/Iy = N1. = 0, the resulting equation is exact. Integrating ]\/I with respect
to 3:, while holding y constant, results in iMz, y) = my + My) . Differentiating with
respect to y, 1% = a: + h’(y) . Setting wy = N, we ﬁnd that h’(y) = 0, and hence
My) 2 0 is acceptable. Therefore the solution is deﬁned implicitly as my 2 6. Note
that if my + 1 = 0, the equation is trivially satisﬁed. 6. Write the given equation as (arc — by)d.7; + (bm — cy)dy. Now 114(93, y) 2 am — by
and N (:13, y) : b1: —— cy. Since Jl/Iy 7E Nm , the differential equation is not exact. 8. Jl/I(.r, y) 2 ea" sin y + By and N(m, y) = —39: + eI sin y. Note that .My 79 NT , and
hence the differential equation is not exact. 10. .M(m,y) = y/m + 6:1; and N(m,y) = ln 3; — 2. Since A/[y 2 NT 2 1/m, the given
equation is exact. Integrating N with respect to y, while holding m constant,
results in QMCU, y) = y 111 z — 2y + Mar) . Differentiating with respect to m, m, =
y/a:+h’(m). Setting w, = 14, we ﬁnd that h’(m) = 6x, and hence Mar) 2 Eng.
Therefore the solution is deﬁned implicitly as 39:2 + y 111 a: — 2y = c. 11. lll(:r,y) = :1: 111 y + my and N(a:,y) = y 111 :1: + my. Note that JVIy yé N3, and
hence the differential equation is not exact. 13. ]\/[(:c,y) = 2a:  y and N(:c,y) 2 2y — :5. Since Ivy 2 Nm 2 —1, the equa—
tion is exact. Integrating JVI with respect to :L', while holding y constant, yields = m2 — cry + My). Now wy = —:L' + h’(y). Equating 1% with N results in
h'(y) = 9y, and hence My) = y2. Thus zb(:L', y) = x2 — my + y2 , and the solution
is given implicitly as .172 :Ey + y2 = C. Invoking the initial condition = 3,
the speciﬁc solution is 3:2 — my—l— y2 = 7. The explicit form of the solution is = (a: + x/ 28 — 3:52 ) / 2. Hence the solution is valid as long as 33:2 g 28. 16. ]\/I(.73, y) = y 62” + ac and N(CL‘, y) = bib egmy. Note that All, = 62'” + 2my 62mg,
and Nm : begmy + 2bcL'y 62”. The given equation is exact, as long as b: 1. In
tegrating N with respect to y, while holding :5 constant, results in 2
5239/2 + Mm). Now differentiating with respect to m, m, : yezmy + 11’ Set—
ting wx = ]\/I, we ﬁnd that h’ = 9:, and hence Mm) = 1'2 / 2. We conclude that
2/)(17, y) = 62“! / ‘2 ~— 1'2 / 2 . Hence the solution is given implicitly as 62‘“; + $2 = c. ...
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This note was uploaded on 01/10/2012 for the course MATH 2233 taught by Professor Binegar during the Fall '08 term at Oklahoma State.
 Fall '08
 BINEGAR

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