Exercise 6

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Exercise 6
First order linear differential equations, many variable scalar functions and integrals
1. Solve the following initial value problems
y 0 + 3t = 2te2t , y(0) = 1
1
y(t) =
−3t2 + 2e2t t − e2t + 3
2
(1)
ty 0 + 2y = t2 − t − 1, y(1) = 0
3t4 − 4t3 − 6t2 + 7
y(t) =
12t2
(3)
ty 0 + 2y = sin t,
(4)
y(π/2) = 1
4 sin(t) − 4t cos(t) +
4t2
y(t) =
(2)
(5)
π2
−4
(6)
2. In the following problems,
(a) Find the solution of the given initial value problem in explicit form.
(b) Plot the graph of the function.
(c) Determine (At least approximately) the interval in which the solution is defined.
2x
,
1 + 2y
(1 + 2y)dy = 2xdx
y0 =
y+y
2
y(0) = −2
(7)
(8)
2
= x +C
(9)
From the initial conditions C = 2
y 2 + y − x2 − 2 = 0
(10)
Reorganising a bit
(y + 1/2)2 − x2 =
9
4
(11)
which are just hyperbolas.
1
0 = xdx + ye−x dy,
y(0) = −1
(12)
xex dx + ydy = 0
y2
x(ex − 1) +
= C
2
1
y(0) = −1 ⇒ C =
2
(13)
(14)
(15)
3. Solve the initial value problem
y0 =
2 cos(2x)
,
3 + 2y
y(0) = −1
(16)
and determine where the solution attains its maximum value.
4. Homogeneous Equations. If the right side of the equation dy/dx = f (x, y) can be expressed
as a function of the ration y/x only, then the equation is said to be homogeneous. Such
equations can always be transformed into separable equations by change of the dependent
variable. This problem illustrates this method.
Consider the equation
dy
y − 4x
=
dx
x−y
(17)
Rewrite the equation as a function of y/x.
Introduce a new dependent variable v so that v = y/x, or y = xv(x).
Express dy/dx in terms of x, v, and dv/dx.
Substitute these expressions in the differential equation and solve for v in terms of x and hence
for y in terms of x.
|y + 2x|3 |y − 2x| = c
(18)
5. Find the volume of the region bounded above by the paraboloid z = x2 + y 2 and below by
the square R : |x| ≤ 1, |y| ≤ 1.
ZZ
V
=
Z
1
Z
1
z(x, y)dxdy =
R
2
2
Z
1
(x + y )dxdy = 4
−1
−1
−1
x2 dx =
8
3
(19)
6. In the following items, sketch the region bounded by the given lines and the regions area as a
double integral and evaluate the integral
(a) The parabolas x = y 2 and x = 2y − y 2
Z 1 Z 2y−y2
1
dy
dy =
3
0
y2
2
(20)
(b) The curves y = ln x and y = 2 ln x and the line x = e, in the first quadrant.
Z e
Z e Z 2 ln x
ln x dx = (x ln x − x)|e1 = 1
dy =
dx
(21)
1
ln x
1
7. Change the Cartesian integral into an equivalent polar integral. Then evaluate the integral
Z
√
1
Z
dx
−1
−
1−x2
√
1−x2
2
dy = π
(1 + x2 + y 2 )2
8. By transforming to cylindrical coordinates, compute the integral
Z
Z √1−y2
1
dy
−1
Z π/2
=
dx
0
xp
x2
+
y 2 dz
Z
0
1
r3 dr =
Z
π/2
=
Z
dθ
−π/2
0
cos θdθ
−π/2
Z
1
Z
rdr
0
r cos θ
rdz =
(22)
0
1
2
(23)
(24)
9. Average values. In polar co-ordinates, the average value of a function over a region R is
given by
Z Z
1
f (r, θ)rdrdθ
Area(R)
R
p
Find the average height of the cone z = x2 + y 2 above the disk x2 + y 2 ≤ a2 in the xy plane.
1
hhi =
πa2
ZZ p
Z 2π Z a
1
1 2πa3
2
2
2
2
x + y rdrdθ =
r
drdθ
=
= a
2
2
πa 0
πa 3
3
R
0
3
(25)