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Math 334
Assignment 5 — Solutions
- Find the solution to the following initial value problems.
(a) y
′′′ − y
′′ − 4 y
′
′ (0) = − 1 , y
′′ (0) = −19.
(b) y ′′′ − 4 y ′′
- 7y ′ − 6 y = 0; y(0) = 1, y ′ (0) = 0, y ′′ (0) = 0.
Solution
(a) The characteristic polynomial for this d.e. is:
C(r) = r
3 − r
2 − 4 r + 4 = (r + 2)(r − 1)(r − 2),
whose roots are r = − 2 , 1 , 2. The general solution to the equation is:
y(x) = c 1 e
− 2 x
x
2 x .
Apply the initial conditions:
y(0) = − 4
y ′ (0) = − 1
y ′′ (0) = − 19
c 1 + c 2 + c 3 = − 4
− 2 c 1 + c 2 + 2c 3 = − 1
4 c 1 + c 2 + 4c 3 = − 19
c 1 = − 2
c 2 = 1
c 3 = − 3
Therefore the solution to the initial value problem is
y(x) = − 2 e
− 2 x
x − 3 e
2 x .
(b) The characteristic polynomial for this d.e. is:
C(r) = r
3 − 4 r
2
2 − 2 r + 3),
whose roots are r = 2, 1 ±
2 i. The general solution to the equation is:
y(x) = c 1 e
2 x
x (c 2 cos
2 x + c 3 sin
2 x).
Apply the initial conditions:
y(0) = 1
y ′ (0) = 0
y
′′ (0) = 0
c 1 + c 2 = 1
2 c 1 + c 2 + 2 1 / 2 c 3 = 0
4 c 1 − c 2 + 2
3 / 2 c 3 = 0
c 1 = 1
c 2 = 0
c 3 = −
Therefore the solution to the initial value problem is
y(x) = e 2 x −
2 e x sin
2 x).
- Find the general solution to the following system of two second order differential equations:
d
2 x
dt^2
− x + y = 0, x +
d
2 y
dt^2
− y = e
3 t .
Solution
Differentiate (1) twice with repect to t to get
x
(iv) − x
′′
′′ = 0.
Use (2) to eliminate y ′′ to get
x
(iv) − x
′′ − x + y = −e
3 t .
Now use (1) to eliminate y from this equation to get
x
(iv) − 2 x
′′ = −e
3 t .
We now have a 4
th order o.d.e. in x only. The characteristic polynomial for the homogeneous equation
is C(r) = r
4 − 2 r
2
. The roots are: r = 0, ±
2, with 0 being a double root. Four linearly independent
solutions to the homogeneous equation are: { 1 , t, e
√ 2 t , e −
√ 2 t }. Since the forcing term e 3 t is not a
solution to the homogeneous equation, we look for a particular solution of the form φp(t) = Ae 3 t .
Plugging φp into the equation leads to A = − 1 /C(3) = − 1 /63. Therefore x(t) is:
x(t) = c 1 + c 2 t + c 3 e
√ 2 t
−
√ 2 t −
e
3 t .
We can now use (1) to solve for y as follows:
y(t) = x(t) − x
′′ (t) = c 1 + c 2 t − c 3 e
√ 2 t − c 4 e
−
√ 2 t −
e
3 t .
The general solution to the system of equations is
x(t) = c 1 + c 2 t + c 3 e
√ 2 t
−
√ 2 t −
e
3 t ,
y(t) = c 1 + c 2 t − c 3 e
√ 2 t − c 4 e
−
√ 2 t −
e
3 t .
- Consider the fourth order differential equation:
y
(iv) (x) − k
2 y
′′ (x) = g(x). (1)
(a) Find the general solution to the homogeneous equation.
(b) Show that a particular solution to Eq. (1) can be written in the form
φp(x) =
k 2
xg(x) dx −
x
k 2
g(x) dx +
e kx
2 k 3
g(x)e
−kx dx −
e −kx
2 k 3
g(x)e
kx dx.
(c) Show that the general solution in part (b) can be rewritten in the form
φp(x) =
∫ (^) x
0
g(s)G(x − s) ds,
where
G(ξ) =
k 3
(sinh kξ − kξ).
(d) Determine the simplest form of the general solution to Eq. (1) in the case g(x) ≡ 1.
Thus, a particular solution is given by
φp(x) =
k 2
xg(x) dx −
x
k 2
g(x) dx +
e kx
2 k 3
g(x)e −kx dx −
e −kx
2 k 3
g(x)e kx dx.
(c) We rewrite the particular solution above as follows:
φp(x) =
k 2
∫ (^) x
0
sg(s) ds −
x
k 2
∫ (^) x
0
g(s) ds +
e
kx
2 k 3
∫ (^) x
0
g(s)e
−ks ds −
e
−kx
2 k 3
∫ (^) x
0
g(s)e
ks ds
∫ (^) x
0
g(s)
s
k^2
x
k^2
e
kx e
−ks
2 k^3
e
−kx e
ks
2 k^3
ds
x
0
g(s)
x − s
k^2
e
k(x−s) − e
−k(x−s)
2 k^3
ds
∫ (^) x
0
g(s)G(x − s) ds,
where
G(ξ) =
e
kξ − e
−kξ
2 k 3
ξ
k 2
k 3
(sinh kξ − kξ).
(d) If g(x) ≡ 1, the particular solution becomes
φp(x) =
∫ (^) x
0
G(x − s) ds =
∫ (^) x
0
G(ξ) dξ =
k 3
∫ (^) x
0
(sinh kξ − kξ) dξ
k 3
cosh kξ
k
kξ
2
x
0
k 4 (cosh kx − 1) −
x
2
2 k 2
The general solution is
y(x) = φh(x) + φp(x) = c 1 + c 2 x + c 3 e
kx
−kx
k 4
(cosh kx − 1) −
x 2
2 k 2
This can be simplified:
y(x) = c 1 + c 2 x + c 3 e
kx
−kx
k 4
e kx − e −kx
x 2
2 k 2
c 1 −
k 4
c 3 +
2 k 4
e
kx
c 4 −
2 k 4
e
−kx −
x 2
2 k 2
= b 1 + b 2 x + b 3 e
kx
−kx −
x
2
2 k^2
Thus, the simplest form for the general solution is
y(x) = b 1 + b 2 x + b 3 e kx
x
2
2 k 2
