Chem 251 Final Solutions
1. Find the solution to the ODE y 0 =
−(2xy+cos x)
.
x2
Solution
Rewriting the ODE as
(2xy + cos x) dx + x2 dy = 0
we see:
δ
δ 2
(2xy + cos x) = 2x =
x
δy
δx
that it is exact.
R
It’s solution is thus 0 = u(x, y) = (2xy + cos x) dx = x2 y + sin x + g(y) where
δ
x2 = δy
u(x, y) = x2 +0+g 0 (y). So g(y) = c. Thus 0 = u(x, y) = x2 y+sin x+c
x+c
is the general solution. Solving for y, we can write this as y = − sin
.
x2
2. Solve the initial value problem y 00 − 2y 0 + 10y = 0, y(0) = 2, y 0 (0) = 5.
Solution
The characteristic polynomial is 0 = γ 2 − 2γ + 10 = (γ − 1)2 + 9 so has
complex roots γ = 1 ± 3i. The general solution to the ODE thus
y = et (c1 cos 3t + c2 sin 3t).
Using the first initial value we find that 2 = y(0) = c1 and using the second
in y 0 = et (−3c1 sin 3t + 3c2 cos 3t) + y we find that 5 = y 0 (0) = (3c2 + y(0)) =
3c2 + 2 yielding that c2 = 1. Thus the solution to the IVP is
y = et (2 cos 3t + sin 3t).
3. Given that the ODE y 00 + y 0 = 0 has the general solution yh = c1 + c2 e−x
find the solution to y 00 + y 0 = sin x.
(Note that I had a mistake in the question, saying that yh = c1 + c2 ex .
This only mattered if you used the method of variation of parameters.
You didn’t lose marks for mistakes due to this. )
Solution
Using the method of undetermined coefficients we predict a solution of the
form yp = A sin x + B cos x. Thus yp0 = A cos x − B sin x and yp00 = −A sin x −
B cos x. Plugging these into the original ODE we get the equations
sin x = yp00 + yp0 = −A sin x − B cos x + A cos x − B sin x
Which has a solution for 1 = −A − B and 0 = A − B. Thus A = B = −1/2.
The general solution to the ODE is thus
y = yh + yp = c1 + c2 e−x −
1
1
cos x − sin x.
2
2
4. Find the first solution to the following, using the Power Series Method or
the Frobenius Method: x2 y 00 + x(2x − 1)y 0 + (x + 1)y = 0.
Solution
As (2x − 1)/x is not analytic at 0, we use the Frobenius Method instead of the
Power Series Method. As (2x − 1) and (x + 1) are analytic at P
0 the Theorem
m+r
of Frobenius tells us that we have a solution of the form y = ∞
m=0 a0 x
for non-zero a0 . Plugging this and its derivatives into the original ODE gives
0 = xm+r
∞
X
am (m
m=0
∞
X
2am−1 (m + r)(m + r − 1) −
+
+
+ r)(m + r − 1)
m=1
∞
X
∞
X
am (m + r)
m=0
∞
X
am−1 +
m=1
am
(1)
m=0
Setting m = 0 and equating the coefficients of xr we get that
0 = a0 (r)(r − 1) − a0 (r) + a0 = a0 (r − 1)2
As a0 6= 0 this yields r = 1. Putting r = 1 into Equation (1) we get by
equating the coefficients of xm+1 that for m ≥ 1 that
am =
−(2m + 1)
am−1 .
m2
Solving for the ai we get first solution
y = a0 x(
∞
X
(−1)m
m=0
xm · (3 · 5 · 7 · · · · · 2m + 1)
)
m! · m!
5. Find the Laplace transform of f (t) = eπt + (t2 + 2)u(t − 1).
Solution
1
+ L ((t2 + 2)u(t − 1)).
L (f (t)) = L (eπt ) + L ((t2 + 2)u(t − 1)) = s−π
2
To evaluate the second part we write t + 2 as a function of v = t − 1:
t2 + 2 = (v + 1)2 + 2 = v 2 + 2v + 3 = (t − 1)2 + 2(t − 1) + 3. Thus by the second
shifting theorem L ((t2 + 2)u(t − 1)) = e−s L (t2 + 2t + 3) = e−s ( s23 + s22 + 3s ).
Altogether, we have
L (f (t)) =
1
2
2
3
+ e−s ( 3 + 2 + ).
s−π
s
s
s
6. Solve the following using the Laplace Transform Method: y 00 −9y = 10e−2t ,
y(0) = y 0 (0) = 0.
Solution
Taking the transforms of both sides of the equation, we have
10
s+2
=
L (10e−t ) = L (y 00 − 9y) = L (y 00 ) − 9L (y)
=
s2 Y − sy(0) − y 0 (0) − 9Y
=
s2 Y + 9Y = Y (s2 − 9)
Solving for Y gives Y =
with partial fractions:
10
(s+2)(s2 −9)
=
10
s
(
+ 2)(s − 3)(s + 3) which we split
10
A
B
C
=
+
+
(s + 2)(s − 3)(s + 3)
s+2
s−3
s+3
=
A(s2 − 9) + B(s2 + 5s + 6) + C(s2 − s − 6)
.
(s + 2)(s − 3)(s + 3)
So 10 = A(s2 − 9) + B(s2 + 5s + 6) + C(s2 − s − 6) which yields the system
A + B + C = 0, 5B − C = 0, and −9A + 6B − 6C = 10. Solving this gives
A = −2 and B = 1/3 and C = 5/3.
Thus our solution is
y = L −1 (Y )
−2
1/3
5/3
) + L −1 (
) + L −1 (
)
s+1
s−3
s+3
=
L −1 (
=
−2e−2t + (1/3)e3t + (5/3)e−3t
A
(Observe that splitting the fraction into s+2
and s2B−9 you would get the
solution
y = −2e−2t + 2 cosh 3t − 4/3 sinh 3t.
It is the same solution.)