MA3264, Tutorial 1, Solutions Lin Meixia, Department of Mathematics
1. (a) x(x + 1)y 1)y = 1 Solution:
y =
1 1 = x(x + 1) x
− x +1 1 ⇒ y = ln |x| − ln |x + 1| + C = ln | x +x 1 | + C.
Substitute (1, (1, 12 ),
1 1 = ln + C 2 2
⇒ C = 21 − ln 12 = 21 + lnln 2. ⇒
The solution is
| | − ln |x + 1| + C = ln | x +x 1 | + 21 + lnln 2.
y = ln x 2.
sec(x cos(5x) sec(x) y = cos(5x
Solution:
1 y = cos(x cos(x) cos( cos(55x) = [cos(6x [cos(6x) + cos(4x cos(4x)] 2
1 ⇒ y = 21 [ 16 sin(6x sin(6x) + sin(4x sin(4x)] + C. 4
Substitute (1, (1, 12 ), 1 1 1 = sin(6) + sin(4) + C 2 12 8
⇒ C = 21 − 121 sin(6) − 18 sin(4). ⇒ sin(4).
The solution is y = 3. y = e x
1 1 1 sin(6x sin(6x) + sin(4x sin(4x) + 12 8 2
sin(4). − 121 sin(6) − 18 sin(4).
y
−3
Solution:
Substitute (1, (1, 12 ),
dy ex = 3y dx e
3y
⇒ e
1 e = e + C, 3 3 2
The solution is
dy = e = ex dx
1 3y 1 e = ex + e 3 3
3 2
⇒ 31 e
3y
= e x + C.
⇒ C = 31 e − e. 3 2
− e or y = 31 ln(3e ln(3e
x
1
+e
3 2
− 3e).
4. (1 + y)y + (1 Solution:
− 2x)y
Substitute (1, 12 ),
The solution is
2
= 0, y > 0
1+y dy = (2x y2
− 1)dx ⇒ − y1 + ln y = x − x + C. 2
−2 + ln 12 = C ⇒ C = −2 + ln 12 = −2 − ln 2. − y1 + ln y = x − x − 2 − ln 2. 2
If we use Graphmatica, we can get the following graph. We can see that we get exactly the
same graphs if we graph these specific functions or feed the differential equations into Graphmatica directly.
2
3
2. Solution: Since y = y, we have y = y = y and so on, then all of the derivatives of y are equal to each other. So since y(0) = 1, we have y(0) = y (0) = y (0) = = 1. Since
·· ·
y = a0 + a1 x + a2 x2 +
·· · ,
we have y(0) = a 0 = 1,
y (0) = a1 = 1 y (0) = 2a2 = 1,
y (0) = 6a3 = 1,
···
thus a n = n1! , that is y = e x. This shows that this is the ONLY solution, PROVIDED that you can use the equation itself to evaluate all of the derivatives of y at y(0).
√
For y = 2 y, y(x)
≥ 0, y(0) = 0, this method does’t work. Since
y =
√ 1y y ,
we will be dividing by zero when we try to evaluate y (0) using the equation. [Remember: the chain rule requires all functions concerned to be differentiable at the points where you are evaluating things. But the square root function is NOT differentiable at zero!] We can solve this differential equation by separating the variables, then we get y = x 2. But we can see that y = 0 is also a solution. So we see that in some [luckily rare] cases, a differential equation can have more than one solution, EVEN if the initial conditions are specified.
4
3. Solution: Denote volume as V and surface area as S . Since the volume of a raindrop is proportional to the 3/2 power of its surface area, we have 3
V = aS ,
(1)
2
where a > 0, with no units. This is reasonable because volume has units of cubic metres and area has units of square metres. [Of course, “reasonable doesnt mean that its always exactly true.] From (1) we differentiate with respect to t on both sides, we can get dV 3 dS = a S . dt 2 dt 1
(2)
2
Since the rate of reduction of the volume of a raindrop is proportional to its surface area, we have
− dV = bS, dt
(3)
where b > 0, with units of metres/sec. This is reasonable because evaporation takes place at the surface of the drop and so its rate can be expected to depend on the area. From (2) and (3), we have 3 dS a S = bS, 2 dt namely, dS b = dt. 3a 2S 1
−
2
−
1 2
Integrating from S 0 to zero, we have 0
S 0
T
1 1
2S
dS =
0
2
Thus T =
3b a
− 3ab dt.
S , 0
that is the amount of time it takes for a virga raindrop to evaporate completely is
3b
a
√ S . 0
Suppose the rate of reduction of the volume of a raindrop is proportional to the square of the surface area, i.e. = bS . − dV dt 2
(4)
Combine (3) and (4), we have dS 3
=
− 3ab dt.
2S When we try to integrate this from S 0 to zero, we will get a divergent integral, meaning that the evaporation would take infinite time and the rain would always reach the ground, contrary to the definition of Virga. 2
5
4. Solution: We use polar coordinates.
Figure 1: after time dt, moth fly from A to B Note: ψ is the angle between the radius vector of the moth [pointing outwards] and her velocity. In ABC we can get the following equation
tan ψ = thus
dr dθ = r tan ψ
rdθ , dr
⇒ r = R exp{ tanθ ψ },
where R is the initial distance. If ψ > π2 , (the moth looking towards a candle in front of her), then tanψ < 0, r will get steadily smaller as θ increases. So the unfortunate moth will spiral into the candle. If ψ < π2 , (the moth’s first view of the candle is over her “shoulder”), then tan ψ > 0, r will get larger as θ increases. So the moth will spiral outwards. if ψ = π2 , the moth will fly along a circle until it drops dead from exhaustion or starvation, whichever comes first. If we use Graphmatica, we can get the following graph.
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Figure 2: ψ =
π 2
Figure 3: ψ > π2 , that is tan(ψ) < 0
Figure 4: ψ < π2 , that is tan(ψ) > 0
7
5. Solution: Since a(t) is a ratio, it has no units. The units of L are those of time. We will prove that a(t) 1. We prove by contradiction. If a(t) < 1, we have
≥
RHS = a2 which is a contradiction. a(t)
− a2
2
+ 1 < 0
2
a2
> 2, thus
but LHS
≥ 0,
≥ 1 means the universe is never smaller than a certain minimum size.
Let y = a 2 , then y = ˙ 2aa˙
y˙ ⇒ a = ˙ . 2a
Plug this into the differential equation, we get
y˙ L
2
2
2a
= y
(y) ˙ 2 = y 4y dy . 1 2 9 (y + 2 ) 4
− y2 + 1 ⇒ L
2
⇒ L2 dt = Let y +
1 2
= 32 cosh(x), [since y = a 2
− y2 + 1 ⇒ L2 dt =
dy
y2 + y
−2
= y
− y2 + 1
−
≥ 1, LH S ≥
3 2
and cosh
≥ 1], then we have
3 sinh(x)dx 2 2 dt = 2 3 . = dx t = x L L sinh(x) 2 3 1 3 2 a2 = y = cosh(x) = cosh( t) 2 2 2 L
⇒
⇒
−
− 12 .
The graph of a starts out at a = 1, and then steadily increases at a faster and faster rate. This describes a Universe that begins with non-zero size and then expands. The expansion is not slowing down — instead it is getting faster and faster. If we use Graphmatica to solve the differential equation directly , we will get that a(t) is identically equal to 1. And in fact this is also a solution of the equation. Actually the solution with a(t) identically equal to 1 is not too interesting because it is not stable. To see this using Graphmatica, compare what happens when you put in 0, 1.00000001 as initial data with what happened when you put in 0, 1 . Even a tiny change makes a big difference this is what we mean by unstable.
{
{ }
8
}
Figure 5: our solution
Figure 6: solve this equation using Graphmatica, type dy = sqrt(y2
Figure 7: solve this equation using Graphmatica, type dy = sqrt(y2
9
2
− (2/y ) + 1){0, 1}
2
− (2/y ) + 1){0, 1.00000001}
6. (a)
1 2y 4x 1 + y + 2x
− −
y = Solution:
Let v = 2x + y, then we have v = 2 + y , thus 1 2v 3 v 2= v = (1 + v)dv = 3dx 1+v 1+v 1 v + v 2 = 3x + C 2 1 (2x + y) + (2x + y)2 = 3x + C. 2
− ⇒
−
⇒
⇒ ⇒
6. (b)
x + y + 1 y =
2
x + y + 3
Solution:
Let v = x + y, then we have v = 1 + y , thus v
2
−
2
2
2
2
v + 1 ⇒ v = 2v + 8v + 10 ⇒ v + 6v + 9 dv = 2dx v + 3 v + 6v + 9 v + 4v + 5 2v + 4 dv = 2dx ⇒ 1+
1=
v 2 + 4v + 5 v + ln v 2 + 4v + 5 = 2x + C
⇒ | | ⇒ (x + y) + ln |(x + y)
2
+ 4(x + y) + 5 = 2x + C.
|
6. (c) x + y + 1 + ( x + y
−
Solution:
− 3)y
=0
Let x = X + α and y = Y + β , then we have x + y + 1 = X + Y + (α + β + 1)
−x + y − 3 = −X + Y + (−α + β − 3).
If we choose
α + β + 1 =
−α + β − 3 = 0, we get α = −2 and β = 1. Meanwhile X = x +2 and Y = y − 1. The differential equation becomes Y Y (X + Y ) + ( −X + Y )Y = 0 ⇒ (1 + ) + ( −1 + )Y = 0. X X
Set V =
Y , X
we will have Y = V X + V and 2
dV 1 dV ⇒ 11+−V V dV = dX ⇒ − X 1 + V 2 1 + V 1 ⇒ arctan V − 2 ln(1 + V ) = ln |X | + C ⇒ arctan( y − 1 ) − 1 ln(1 + ( y − 1 ) ) = ln |x + 2| + C.
(1 + V ) + ( 1 + V )(V X + V ) = 0
−
2
2
2
x+2
2
x+2
10
2
2
=
dX X
