Download Quiz 3 solution

MATH 225 / MTHE 225
Quiz 3 - Solution
1. Solve the differential equation: y 00 + 5y 0 + 7 = 0
It should be y 00 + 5y 0 + 7y = 0. But still you can solve it.
Solution: (of the original/uncorrected version)
First rewrite the DE as: y 00 + 5y 0 = −7.
• For yc : Set y 00 + 5y 0 = 0
⇒ Auxiliary equation: m2 + 5m = 0 ⇒ m(m + 5) = 0 ⇒ m1,2 = −5, 0.
⇒ yc = C1 e−5·x + C2 e0·x = C1 e−5x + C2 .
• For yp : Try yp = A Ax (we can not use A in order to avoid the similar type of
terms in between yc and yp ). Therefore, yp0 = A ⇒ yp00 = 0. Now pug all these in
the DE:
5
0 + 5A = −7 ⇒ A = − .
7
5
So, yp = − Ax.
7
5
• Therefore the general solution of the DE is: y = yc + yp = C1 e−5x + C2 − Ax.
7
Solution: (of the corrected version):
2
Auxiliary equation: m + 5m + 7 = 0 ⇒ m1,2 =
−5 ±
(= α ± βi).
The general solution of the given DE is:
"
y = eαx [C1 cos (βx) + C2 sin (βx)] = e
−5
x
2
C1 cos
√
52 − 4 · 1 · 7
−5
3
=
±
i
2·1
2
2
√
√ !
3
x + C2 sin
2
√
3
x
2
!#
.
2. Find the appropriate form of the particular solution yp (but do NOT evaluate the coefficients) of the differential equation: y 00 + 3y 0 + 2y = 1 + e−x + cos (3x).
Solution: We use “undetermined coefficients or trial yp ” here.
• For yc : Set y 00 + 3y 0 + 2y = 0.
⇒ Auxiliary equation: m2 + 3m + 2 = 0 ⇒ (m + 1)(m + 2) = 0 ⇒ m1,2 = −2, −1.
⇒ yc = C1 e−2x + C2 e−x .
• For yp : Try yp = A + Be−x Bxe−x + C cos (3x) + D sin (3x) (we can not use Be−x
in order to avoid the similar type of terms in between yc and yp ).
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3. Solve the differential equation: y 00 + 3y 0 + 2y =
1
.
1 + ex
Solution: Here we use “variation of parameters”. At first notice that the left side of
this DE is exactly same as that of the previous question. So we can just copy the first
portion from there; that is yc = C1 e−2x + C2 e−x . That means the linearly independent
solutions of the homogeneous part are: y1 = e−2x and y2 = e−x .
−2x
−x e
e
= −e−3x + 2e−3x = e−3x
W = −2e−2x −e−x 0
e−x
e−x
e−x W1 = 1
=
0
−
=
−
−e−x 1 + ex
1 + ex
1+ex
−2x
e
e−2x
e−2x
0 W2 = −
0
=
1 =
−2x
−2e
1 + ex
1 + ex
1+ex
e2x
W2
ex
W1
0
=−
and
u
=
=
2
W
1 + ex
W
1 + ex
Z
Z
Z x x
e2x
e (e dx)
0
u1 = u1 dx = −
dx = −
. Now apply u-substitution with u =
x
x
1+e
Z x x1+e
Z
Z e (e dx)
(u − 1) du
1
x
x
1 + e ⇒ du = e dx. So u1 = −
=−
=−
1−
du =
1 + ex
u
u
−(u − ln u) = −(1 + ex ) + ln (1 + ex ).
u01 =
ex
u2 =
dx =
dx. Now apply u-substitution with u = 1+ex ⇒ du = ex dx.
1 + eZx
Z
ex
du
So u2 =
dx
=
= ln u = ln (1 + ex ).
x
1+e
u
Z
u02
Z
Therefore, yp = u1 · y1 + u2 · y2 = e−2x [−(1 + ex ) + ln (1 + ex )] + e−x ln (1 + ex ).
Therefore the general solution of the given DE is:
y = yc + yp = C1 e−2x + C2 e−x + e−2x [−(1 + ex ) + ln (1 + ex )] + e−x ln (1 + ex ).
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