Download Solutions to Assignment 1

MTHE 237
Fall 2015
Solutions to Assignment 1
Problem 1
a) Find the order of the differential equation:
t
d3 y
+ t2 y = cos(t).
dt3
Is the differential equation linear? Is the equation homogeneous?
b) Repeat the above for the following differential equation:
d2 y
dy
+ 3y
= 0.
dt2
dt
c) Consider the equation
y ′′ − 3y ′ + 2y = 0,
By trying a solution of the form y = ert , find the possible r values satisfying the differential equation.
Solution:
a) The order is 3, and the equation is linear. The equation is non-homogeneous.
b) The equation is non-linear (The reason here is that yy ′ is not a linear function of (y, y ′ )). The equation
has order 2. The equation is homogeneous.
′′
c)By substituting y = ert , and noting that y ′ = rert and y = r2 ert , it follows that
ert (r2 − 3r + 2) = 0,
since ert 6= 0, it follows that
(r − 2)(r − 1) = 0,
which implies that r = 1 or r = 2.
Problem 2
Consider the following differential equation
dy
= −ay + b,
dt
where a > 0, b > 0 are positive real-valued numbers.
a) Obtain the general solution of the differential equation.
b) Sketch the particular solutions for three different initial conditions, that is for three different y(0) values.
c) Describe how the solutions change if (i) a increases (ii) b increases. (iii) b = 0 and a increases.
d) Draw a qualitative direction field for the equation
dy
= −5y + 10
dt
for t ≥ 0. In your graph, highlight the particular solution passing through the point (t, y) = (0, 10).
Solution:
a) The equation
dy
= −ay + b
dt
can be written as
dy
= dt
−ay + b
It follows by multiplying boths sides by −a and integration that
ln(|y − b/a|) = −at + C,
for some constant C. Hence, if y > b/a, we have
y(t) = b/a + Ke−at ,
and if y < b/a, we have
y(t) = b/a − Ke−at ,
with K = eC .
b,c,d)If y(0) = y0 , then K = b/a − y0 . Thus,
y(t) = b/a + (y0 − b/a)e−at ,
If a increases, the equilibrium value is approached more rapidly, the equilibrium value is b/a.
Problem 3
According to Newton’s law of cooling, the temperature at time t, u(t), of an object satisfies the following
differential equation:
du
= −k(u − T ),
dt
where T is the constant temperature of the environment and k > 0 is a positive constant. Let the initial
condition be such that u(0) = u0 .
a) Find the temperature of the object as a function of t.
b) Let τ > 0 be the time at which the initial temperature difference between u(t) and T , u0 − T has been
reduced by half, that is
u0 − T
.
u(τ ) − T =
2
Find the relation between k and τ .
Solution:
a) We have
du
= −k(u − T )
dt
with u(0) = u0 . It follows that
ln(|u − T |) = −kt + C.
Hence,
u = Ke−kt + T
for some constant K. Since u(0) = u0 , we have that K = u0 − T . Hence,
u(t) = (u0 − T )e−kt + T
b) We have that
u(τ ) − T =
1
(u0 − T ),
2
hence, e−kτ = 1/2 and
kτ = ln(2)
Problem 4
Consider the following differential equation:
(2x + 4xy)dx + (2x2 + 3y 2 )dy = 0,
defined on x ≥ 0, y > 0
a) Is the equation exact?
b) Obtain a general (implicit) solution to the equation:
c) Find the particular implicit solution which passes through the point (x, y) = (0, 5).
Solution:
a)
M (x, y) = 2x + 4xy
N (x, y) = 2x2 + 3y 2
It follows that
∂
∂
M (x, y) =
N (x, y) = 4x
∂y
∂x
Hence, the equation is exact.
b) Due to exactness, by a theorem that we proved in class, there exists a function F (x, y) such that Fx =
M (x, y) and Fy = N (x, y)
Upon integrating M (x, y) along the x coordinate we obtain:
F (x, y) = x2 + 2x2 y + g(y) + C
for some g(.) which is a function of y only. Now, taking the partial derivative of this term with respect to y
and equating it to N , it follows that g ′ (y) = 3y 2 , which leads to g(y) = y 3 plus some constant term, which
is already accounted for by C. Hence, F (x, y) = x2 + 2x2 y + y 3 + C = 0, is a general solution.
c) To find the particular solution at (1, 1), we substitute the values and obtain that C = −4. As such, the
particular solution is
x2 + 2x2 y + y 3 = 4.
There is a unique particular solution by the existence and uniqueness theorem since
dy/dx = −
3x2 + 4xy
,
2x2 + 3y 2
has continuous partial derivatives in y for y > 0.
Problem 5
Solve the differential equation xdx + ydy = 0 with initial condition y(1) = 1.
Solution:
∂
∂
M (x, y) = 0 = ∂x
N (x, y).
This is an exact differential equation since with M (x, y) = x, N (x, y) = y, ∂y
Rx
Find F (x, y) =
M (s, y)ds + K(y) = x2 + K(y). Taking the partial derivative with respect to y and
equaling to N leads to K(y) = y 2 /2 + c. As a result F (x, y) = (1/2)(x2 + y 2 ) + c = 0 is a solution. y(1) = 1
implies that c = 2 (Note that this is the only solution by the existence and uniqueness theorem since when
|x| > 0, |y| > 0, −x/y is continuous with continuous partial derivatives).
Problem 6
dy
= f (x, y), where f (x, y) is a continuous function and f (x, y) has
Consider a differential equation dx
continuous partial derivatives in y, on D ⊂ R2 which is a closed and bounded set. Consider two different solutions corresponding to two different initial conditions (x0 , y0 ) and (x0 , ỹ0 ) where (x0 , y0 ) ∈ D and
(x0 , ỹ0 ) ∈ D. Suppose, furthermore that the differential equation is defined only on D.
Can these two solutions intersect at any point in D?
Solution:
We will use contradiction to arrive at the correct conclusion. Suppose that there exists a point (x̃, ỹ) ∈ D
such that the two solutions pass through this point and they start to behave differently after this point. By
the existence and uniqueness theorem, this implies that there is a unique solution in a small neighborhood
of this point. Then, there cannot exist such a point. The solutions are then either identical everywhere; or
do not intersect at all.
Problem 7
Show that given a function F , if
∂
∂
∂x ( ∂y F (x, y))
and
∂
∂
∂y ( ∂x F (x, y))
are continuous, they must be equal.
Solution:
This result is known as Clairaut’s Theorem. Let a, b, c, d ∈ R, a < b and c < d. Let Fx =
∂
∂
∂
∂
∂x ( ∂y F ) = Fyx and ∂y ( ∂x F ) = Fyx .
∂
∂x F ,
Fy =
∂
∂y F ,
Now, by the fundamental theorem of calculus
Z
a
b
Z
d
Fxy dydx =
Z
a
c
b
Z
c
d
Fxy dy dx
b
Fx (x, d) − Fx (x, c) dx
=
a
= F (b, d) − F (b, c) − F (a, d) − F (a, c)
Z
(1)
RdRb
A parallel argument applies for the integration c a Fxy dydx leading to
Z dZ b
Fyx dxdy = F (b, d) − F (a, d) − F (b, c) − F (a, c)
c
a
Now, these two imply that, by continuity, they have to be equal inside the integration as well. Suppose not.
Then, for some ǫ > 0, there exists a point (x0 , y0 ) where
|Fyx (x0 , y0 ) − Fxy (x0 , y0 )| ≥ ǫ.
But then, since the functions are continuous, there exists a neighborhood of this point (say (x0 − δ, x0 + δ) ×
(y0 − δ, y0R+ δ) for some δR > 0) where the functions are different by at least ǫ/2. Then, the integral of the
functions Fxy dydx and Fyx dxdy on this neighborhood will be strictly different. Hence, a contradiction.
It must be that Fxy = Fyx , if they are continuous.
Some additional problems:
Problem 8
Consider the family of parabolas defined by
y = kx2 ,
where k is an arbitrary constant. Find the family of curves which intersects the given family of parabolas
orthogonally at each point. You are asked to find an expression in the form F (x, y) = 0 (this may include a
constant term) such that the family of such curves is orthogonal to the family of parabolas for every k.
Hint: Use the fact that a line with slope m at a point P , is orthogonal to the line with slope −1/m passing
through the same point.
Solution:
We have that the solution should satisfy
−1
−1
1
dy
=
=
= − x/y,
dx
2kx
2(y/x2 )x
2
Hence,
xdx + 2ydy = 0
This equation is exact and the solution is given by the ellipsoid:
x2 + 2y 2 = c,
for an arbitrary constant c.
Problem 9
Obtain the general solution to the differential equation
(3x2 + 4xy)dx + (2x2 + 3y 2 )dy = 0,
for x > 0, y > 0.
Solution:
∂
This is an exact differential equation since with M (x, y) = 3x2 + 4xy, N (x, y) = 2x2 + 3y 2 , ∂y
M (x, y) =
∂
4x = ∂x N (x, y).
Rx
Find F (x, y) =
M (s, y)ds + K(y) = x3 + 2x2 y + K(y). Taking the partial derivative with respect to y
and equaling to N leads to K(y) = y 3 + c. As a result F (x, y) = x3 + 2x2 y + y 3 = K is the general solution.