Download Quiz 6 Solution: Differential Equations and Power Functions and more Exercises Calculus in PDF only on Docsity!
Answer Key for Quiz 6 (section B)
- If y = Cxn^ for some constants C and n then
dy dx
= C
nxn−^1
and
d^2 y dx^2
= Cn(n − 1)xn−^2
Plugging y, dydx and d
(^2) y dx^2 into the differential equation
x^2 d^2 y dx^2
we get
0 = x^2
Cn(n − 1)xn−^2
Cnxn−^1
- 2Cxn = Cn(n − 1)xn^ + 4Cnxn^ + 2Cxn = Cxn^ [n(n − 1) + 4n + 2] = Cxn^
n^2 − n + 4n + 2
= Cxn^
n^2 + 3n + 2
= Cxn(n + 2)(n + 1).
Since xn^ only equals zero at x = 0, there are only three ways to make the right side identically zero: we can take C = 0, or n = −1, or n = −2. This shows that any function of the form y = Cx or y = (^) xC 2 is a solution of the differential equation, but no other power function could be. In fact every solution of the differential equation has the form
y =
A
x
B
x^2
for some constants A and B.
- Method 1: multiply the differential equation by e^3 t^ to get
e^3 t^ dx dt
This makes the left side into the derivative of xe^3 t:
d dt
xe^3 t^ = 24e^3 t.
Therefore, if we integrate both sides with respect to t we get
∫ 24 e^3 t^ dt =
d dt xe^3 t
dt = xe^3 t.
We can use the substitution u = 3t to do the integral on the left: du = 3 dt, so we have
xe^3 t^ =
24 e^3 t^ dt = 8
e^3 t^3 dt = 8
eu^ du = 8eu^ + C = 8e^3 t^ + C.
We can solve this for x by multiplying through by e−^3 t:
x = 8 + Ce−^3 t.
Now plug in x = 12 and t = 0 to get
12 = 8 + Ce^0 = 8 + C, so C = 4 and we finally have x = 8 + 4e−^3 t.
Method 2: rearrange the differential equation to
dx dt
= 24 − 3 x and then to dt =
dx 24 − 3 x
To make the right side slightly nicer, multiply through by −3 to get
− 3 dt =
− 3 dx (−3)(x − 8)
dx x − 8
Then integrate both sides:
∫ − 3 dt =
dx x − 8 =⇒ − 3 t + C = ln |x − 8 |.
We have several options now.
Method 2a: plug in x = 12 and t = 0 to get
0 + C = ln | 12 − 8 | = ln 4, so C = ln 4.
This also tells us that we can remove the absolute values on x − 8; since it is positive at t = 0 it must stay positive, otherwise ln |x − 8 | would become infinite before x − 8 could become negative. So now we know
ln(x − 8) = − 3 t + ln 4,
and we can solve for x by exponentiating both sides:
eln(x−8)^ = e−^3 t+ln 4^ = e−^3 t^ eln 4^ which simplifies to x − 8 = 4e−^3 t.
So we get x = 8 + 4e−^3 t^ as before.
Method 2b: do the exponentiation step before plugging in. This gives
eln^ |x−^8 |^ = e−^3 t+C^ = e−^3 t^ eC^ which simplifies to |x − 8 | = Ke−^3 t,
where K = eC^ is a positive arbitrary constant. Plugging in x = 12 and t = 0 we now get
| 12 − 8 | = Ke^0 = K, so K = 4 and therefore |x − 8 | = 4e−^3 t.
This implies that either x − 8 = 4e−^3 t^ or x − 8 = − 4 e−^3 t.
But only the former can be right because in the latter x doesn’t equal 12 when t = 0. So once again we get x = 8 + 4e−^3 t.
Method 2c: as in method 2b, but at the stage where we rename eC^ as K, allow K to be a completely arbitrary constant and use that freedom to remove the absolute values. This gives x − 8 = Ke−^3 t, and when we plug in x = 12 and t = 0 we get K = 4 as before.
