DIFFERENTIAL EQUATIONS
439
46.59
Solve y" — 5y' + 4y = sin3jc.
A general solution of the homogeneous equation y" - 5y' +4y =0 was found in Problem 46.50; it is
C^e* + C 2 e
4 *. Let us guess that a particular solution of the given equation is y = A sin 3x + B cos 3x.
(An alternative to guessing is provided by Problem 46.65.) Then y' = 3A cos3x — 38 sin3* and
y" = -9A sin 3x - 9B cos 3x. So sin 3x = (-9 A sin 3x - 9B cos 3*) - 5(3 A cos 3* - 38 sin 3x) + 4(A sin
B cos 3*). Equating coefficients of sin 3* and cos 3x, we get -9A + \5B + 4A = 1 and -9B - 15 A + 4B = 0,
or, more simply, —5/1 + 158 = 1 and — 15 A — 5.8 = 0. Hence, solving simultaneously, we get —50/1 = 1,
and
sin3x) + C ie * + C 2 e
4 *.
Therefore,
(3 cos 3* - sinSx). The general solution is
(3 cos 3x -
y" + y = tan x
(1)
The general solution of y" + y = 0 is obtained by solving the auxiliary equation u
2 + 1 = 0. The roots
are ±i. Hence, the general solution consists of all linear combinations of y 1 = cosx and y 2 = sinx.
Calculate the Wronskian of y 1 and y 2 :
46.65
Solve by the method of variation of parameters,
46.64
Solve y" + Av = B, where A > 0.
The general solution of the corresponding homogeneous equation is y g = C l cos ^^Ax + C 2 sin V~Ax, by
Problem 46.55. A particular solution is y p = B/A. Hence, the general solution is BIA + C l cosVAx +
C 2 sin VAx.
46.63
Find a general solution of y" — y' — 6y = 2(sin 4x + cos 4x).
A general solution of y"-y'—6y = 0 is obtained by solved the auxiliary equation u — M —6 = 0. Since
M2 — M - 6 = (M — 3)(« + 2), the roots are u = 3 and u = -2. Hence, the general solution is C^3* +
C 2 e~
2 '. Now we try y = A sin 4x + B cos 4x as a particular solution of the given equation. Then y' =
4A cos 4x -4B sin 4x, and y" = -16A sin 4x - 168 cos 4x. Thus, y" - y' -6y = (4B - 22,4) sin 4* - (4A +
22B)cos4x = 2sin4;e + 2cos4x. Hence, 48-22/1 = 2 and -22B-4A = 2. Solving simultaneously,
we get
and
(13 sin 4* + 18cos4jc), and the general solution is
So
(13 sin 4x + 18 cos 4x) + C,e
3 * + C,e~
2 '.
46.60
Solve y" + 2y' + 2y = e
3 *.
The general solution of the corresponding homogeneous equation was found in Problem 46.53: y g =
(C, cos x + C 2 sin x)e~*. Let us try y = Ae** as a particular solution of the given equation. Then, y'=
3Ae
3 ' and v" = 9Ae
3 *. So 9Ae
3 ' + 6Ae
3 * + 2Ae
3 " = e
3 '. Hence, 15/1 = 1. A=&. Thus, a particular solution is
and the general solution is
(Q cos x + C 2 sin x)e ".
46.61
Solve y" + y'-12y = xe*.
A general solution of the corresponding homogeneous equation is CjC
3 * + C 2 e **, since 3 and —4 are
the roots of the auxiliary equation u
2 + u - 12 = 0. Let us try y = (A + Bx)e* as a particular solution
of the given equation. Then y' = (A + B + Bx)e", and y" = (A + 2B + Bx)e*. Thus, (A + 28 + Bx)e* +
(A + B + Bx)e* - \2(A + Bx)e* = xe'. Hence, (A + 2B + Bx) + (A + B + Bx) - U(A + Bx) = x. Equating
the coefficients of x as well as the constant coefficients, we get —10B = 1 and —10 A + 38 = 0. Therefore,
and the general solution is
Thus,
and
46.62
Find a general solution of
y" + 9y = e' cos 2x
(1)
A general solution of the corresponding homogeneous equation was found in Problem 46.55: y g =
C 1 cos 3x + C 2 sin 3x.
Let us try
y = Ae* cos 2x + Be" sin 2x
as a particular solution of (1).
Then
y' = A(e* cos 2x - 2e* sin 2x) + B(e* sin 2x + 2e* cos 2x) = (A + 2B)e" cos 2x + (B - 2A)e* sin 2x, and v" =
(45 - 3A)e* cos 2x - (3B + 4A)e" sin 2x. Thus, e* cos 2x = (4B - 3A)e' cos 2x - (3B + 4A)e* sin 2x +
9(Ae" cos2x + Be" sin 2*). Equating coefficients of e* cos2x and e*sin2x, we get 4B — 3A + 9A = 1
and -(3,8 + 4,4)+ 98 = 0, which, when simplified, become 48+6/1 = 1 and 68-4,4 = 0. Solving
simultaneously yields
and
Hence,
(3e* cos 2x + 2e* sin 2x). The general solution
(3e* cos 2x + 2e* sin 2x) + C 1 cos 3x + C 2 sin 3x.