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Unformatted text preview: 148 4.3 Chapter 4. Higher Order Linear Equations 2. The general solution of the homogeneous equation is
yC : clet + 026% + 03 cos t + C4 sin t. Let 91(t) 2 3t and 92(75) : cos t. By inspection, we ﬁnd that Y1(t) 2 —3t. Since
g2(t) is a solution of the homogeneous equation, set 160:) : t(A cos 75+ Bsin t).
Substitution into the given ODE and comparing the coefﬁcients of similar term
results in A = 0 and B = —1/4 . Hence the general solution of the nonhomogeneous
problem is t
y(t) = yc(t) — 3t — 1 sin 15. 3. The characteristic equation corresponding to the homogeneous problem can be
written as (r + 1)(7“2 + 1) = O. The solution of the homogeneous equation is yc =
cle—t + Cg cos 15+ 123 sin t. Let gl(t) = e’t and 92(t) : 4t. Since g1(t) is a solution
of the homogeneous equation, set Yl (t) = Ate’t. Substitution into the ODE results in A : 1/2. Now let Y2(t) : Bt + C. We ﬁnd that B = —C : 4. Hence the general
solution of the nonhornogeneous problem is y(t) : yc(t) + te’t/2 + 4(t — 1) . 4. The characteristic equation corresponding to the homogeneous problem can
be written as r(7" + 1)(r — 1) = 0. The solution of the homogeneous equation is
yc = c1 + Cget + 0364. Since g(t) = 2 sin 75 is not a solution of the homogeneous
problem, we can set Y(t) : A cos t + B sin t. Substitution into the ODE results in A 2 1 and B = 0. Thus the general solution is y(t) : 01 + C26t —— eye—t + cos t. 6. The characteristic equation corresponding to the homogeneous problem can be
written as (T2 + 1)2 : 0. It follows that ye = cl cos t + 02 sin t + 75(03 cos 75+ C4 sin t). Since g(t) is not a solution of the homogeneous problem, set Y(t) = A + B cos 215 +
Csin 2t. Substitution into the ODE results in A = 3, B 2 1/9, C = 0. Thus the
general solution is y(t) = yc(t) + 3 + 5 cos 2t. 7. The characteristic equation corresponding to the homogeneous problem can be
written as r3(r3 + 1) = 0. Thus the homogeneous solution is yc = 01 + Cg t + C3752 + C4e_t + 6”2 [C5 cos(\/37t/2) + 05 sin(\/3—t/2)]. Note the g(t) = t is a solution of the homogenous problem. Consider a particular
solution of the form Y(t) : t3(At + B). Substitution into the ODE gives us that
A = 1/24 and B 2 0. Thus the general solution is = yc(t) + 254/24. 8. The characteristic equation corresponding to the homogeneous problem can be
written as T3 (7“ + 1) : 0. Hence the homogeneous solution is yC = 61 + C2 t + 03132 + c4e*t. 154 Chapter 4. Higher Order Linear Equations The solution of the system of equations (10) is I _ tW1(t) _
I tW (t) —t
u2(t)— Wit) —te /2.
ago iii/ﬁg) meet/2 Hence u1(t) : —t2/2, uﬂt) = —e_t(t + 1)/2, U305) : et(t — 1)/2. The particular
solution becomes Y(t) = —t2/2 — (t + 1)/2 + (t —1)/2 = —t2/2 — 1. The constant
is a solution of the homogeneous equation, therefore the general solution is ' = c1 + aget + C3€Tt i t2/2. 3. From Problem 13 in Section 4.27 yc(t) = (:1 e": + Cget + 0362‘. The Wronskian is evaluated as W(e_t7 et, (3%) : 6 6%. Now compute the three determinants 0 6t eZt
W1(t) = 0 6t 262t i e3t
1 et 4e2t e't t 0 : ’CTt t 0 : 2
64 t 1 Hence = (aw/6, = ~e3t/2, = egt/B. Therefore the particular so—
lution can be expressed as Y(t) = 6—t [ea/30] — et [em/6] + (22’: haw/6] : 653/30. 6. From Problem 22 in Section 4.27
116(25): 01 cos 15+ 02 sint + t [Cg cos t + C4 sin t]. The Wronskian is evaluated as W(cos t, sin t,t cos t,t sin t) 2 4. NOW compute
the four auxiliary determinants 0 sin t t cos t t sin t
W1“): 0 cost cost~tsint sint—l—tcost : 0 —sint —2sint~tcost 2cost—tsint 1 —cost —3cost+tsint e3sint—tcost = —2sint + 2t cos t, Chapter 4. Higher Order Linear Equations 9. Based on Problem 4 , u1’(t) : sec t, ugl (t) = —17 = — tan t. The particular
solution can be expressed as Y(t) : [U1 + cos t [U2(t) + sin t [U3(t)]. That is7 Y(t) : 1n Isec(t) + tan(t) — t cos t+ sin t in c0s(t) .
Hence the general solution of the initial value problem is
y(t) : 01 + C2 cos t+ C3 sin t + 1n sec(t) + tan(t)1 t cos t+ sin t in cos(t). Invoking the initial conditions, we require that 01 + Cg : 2, 03 = 1, —Cg = —2.
Therefore y(t) = 2 cos t + sin 25+ in lsec(t) + tan(t) —t cos t+ sin 75 ln cos(t) ta
M '0 0.: 11'4‘ as as 1 1'2 14
t 10. From Problem 6, y(t) = 01 cos 23+ 02 sin t + 0315 cos t + C4t sin t  t2 sin 15/8.
In order to satisfy the initial conditions, we require that 01 = 2 , 02 + 03 = 0 , c1 +
204 = —1,*3/4 — (:2 — 363 = 1. Therefore y(t) = 200s t—l— [7sint — 7t cos t+4t sin t—t2sin t] /8.
1D 8 5 12. From Problem 8, the general solution of the initial value problem is ya) : cl + C2€t + C36" +111csc(t) + cot(t)l + 81: t 6—: t
+—/ 6’3 csc(s)ds + —/ es csc(s)ds.
2 to 2 to ...
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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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