Sample Solutions of Midterm for MATH3270A
a
Note: Any problems about the sample solutions, please email Mr.Xiao Yao (yxiao
math.cuhk.edu.hk) directly.
October 21,2013
1.Solve the following initial value problems:
(i)(1 + x2 )dx − x sin tdt = 0,
(ii)ty 0 + 2y = 4t2 ,
x(0) = 1.
y(1) = 2
π
2
(iii)y 0 = (cos( 1+y ))10 (t2 + y 2 )5 ,
y(0) = 0
(a)Find the solution to the given value problem in explicit form in each case.
(b)Determine the interval in which the solution is defined.
Answer:
(i)This is a seperable equation:
1 + x2
dx = sin tdt
x
1
1
ln |x| − 0 + x2 −
= − cos t + 1
2
2
1
3
ln |x| + x2 = − cos t +
2
2
Then ln |x| + 12 x2 ≥
1
2
> 0,and since LHS is increasing with the value of |x|,this shows that x is
always away from 0.Of course the solution of an ODE is continuous,x(0) = 1 > 0 and thus x is
kept above the x-axis all along,i.e. x > 0. hence the solution is
1
3
ln x + x2 = − cos t +
2
2
And the interval is R in which the solution is defined.
1
2
(ii)The integrator is e
R
2
dt
t
= t2 , hence :
(t2 y)0 = 4t3
t2 y − 2 = t4 − 1
1
y = t2 + 2
t
And the interval is (0, +∞) in which the solution is defined.
(iii)Obviously y = 0 is a solution.Let RHS= f (t, y),Obvious f (t, y) and
∂
f (t, y)
∂y
is continuous
at (0, 0),thus by well-posedness y = 0 is the unique solution.And the interval is R in which the
solution is defined.
Remark:(i)The integral of
R 1
x
is ln |x| when we are not sure about x is positive and nega-
tive.And we need some further observations that x > 0 holds true.
(ii)Many students make the same mistakes in quiz 2,a interval is a set of real numbers with
the property that any number that lies between two numbers in the set is also included in the
set(Ref:Wikipedia).For example (0, 1) is a interval while (0, 1) ∪ (1, 2) is not.
(iii)The full version of existence and uniqueness theorem is below:
For the equation:y 0 = f (t, y), y(t0 ) = y0 ,if f and ∂f /∂y are continuous in a rectangle :|t − t0 | ≤
a, |y − y0 | ≤ b,then there is some interval |t − t0 | ≤ h ≤ a in which there exists a unique solution
y = φ(t) of the IVP.
Please pay special attention to the condition that it requires the continuity of f and ∂f /∂y on a
2-D rectangle which includes the initial point, not a interval of t or y.
2.Consider the following linear second order differential equation
y 00 + p(t)y 0 + q(t)y = 0
on I
and p(t) and q(t) are continuous on I.
(a)State and prove the Abel’s formula for two arbitrary solutions to this equation on I.
(b)Can y(t) = t3 be a solution to this equation on I?Why?.
Answer:
3
(a)It is just directly resembling from the textbook.
(b)Method 1 :
Suppose y2 is another solution of the equation then:
W (t3 , y2 )(t) = t3 y20 − 3t2 y2
= t2 (y20 t − 3y2 )
Since 0 ∈I.Hence W (t3 , y2 )(0) = 0 on I.Since y2 is arbitrary solution of the equation,t3 can’t
compose a set of fundamental solutions with y2 .Contraditcion,thus t3 can’t be a solution of the
equation.
Method 2 :
Suppose t3 is a solution,substitute it into the equation:
6t + 3t2 p(t) + t3 q(t) = 0
1
2
p(t) = − tq(t) −
3
t
Since q(t) is continuous on the interval and 0 ∈ I,thus p(t) is not continuous at 0,contradiction.
3.Find the value of the constant of α so that
(2xy − y 2 ey )dy + αy 2 dx = 0
is exact,and then solve the equation for this α.
Answer:
Mx = 2y,
Ny = 2αy
Mx = Ny ⇒ α = 1
2
d(xy −
Z
ey y 2 dy) = 0
xy 2 − y 2 ey + 2yey − 2ey = C
Where C is arbitrary constant.
4
4.Find the general solutions to the following ODE:
(i)y 00 + 6y 0 + 9y = t−2 e−3t ,
t>0
(ii)y 00 + 6y 0 + 9y = t2 e−3t ,
t>0
(iii)y 00 − 2y 0 + y = 2et
Answer:
(i)The characteristic equation of the corresponding homogeneous equations is:
r2 + 6r + 9 = 0 ⇒ r = −3
Assume the other solution is y2 (t) = e−3t u(t),substitute it into the equation then we get:u(t) =
t.Thus the general solution to the homogeneous equation is
y(t) = c1 e−3t + c2 te−3t ,
c1 , c2 ∈ R
The Wronskian is W (y1 , y2 )(t) = e−6t ,so the particular solution is:
Z t −3s −2 −3s
se−3s s−2 e−3s
e s e
−3t
Y (t) = −e
ds + te
ds
−6s
e
e−6s
t0
t0
Z t
Z t
1
1
ds + te−3t
ds
= −e−3t
t0 s2
t0 s
= −e−3t ln t − e−3t − e−3t ln t0 + e−3t t/t0
−3t
Z t
Hence the general solution to the nonhomogeneous equation is:
y(t) = c1 e−3t + c2 te−3t + e−3t ln t,
c1 , c2 ∈ R
(ii)General solution to the homogeneous one is :
y(t) = c1 e−3t + c2 te−3t ,
c1 , c2 ∈ R
Then the particular solution is
Y (t) = −e
−3t
Z t
t0
−3t
= −e
Z t
Z t −3s 2 −3s
se−3s s2 e−3s
e se
−3t
ds
+
te
ds
e−6s
e−6s
t0
3
s ds + te
−3t
t0
= −e−3t ln t − e−3t − e
1 4 −3t
=
te
12
Z t
t0
−3t
s2 ds
ln t0 + e−3t t/t0
5
Then the general solution to the nonhomogeneous equation is :
y(t) = c1 e−3t + c2 te−3t +
1 4 −3t
te ,
12
c1 , c2 ∈ R
(iii)General solution to the homogeneous one is :
y(t) = c1 et + c2 tet ,
c1 , c2 ∈ R
The Wronkian is W (y1 , y2 ) = e2t ,thus the particular solution is :
Z t s s
e 2e
ses 2es
t
ds + te
ds
Y (t) = −e
2s
e
t0 e2s
t0
= −et (t2 − t20 ) + 2tet (t − t0 )
t
Z t
The general solution to the nonhomogeneous equation:
y(t) = c1 et + c2 tet + t2 et ,
c1 , c2 ∈ R
5.Consider the equation
dy
= y(y − 1)(y − 3)
dt
(a)Is this an autonomous equation?
(b)Determine all the critical points.
(c)Are the critical points asymptotically stable or unstable?Why?
(d)Draw the phase line.
(e)Sketch several graphs of solutions in ty-plane and indicate the monotonicity and convexity of
the trajectories.
Answer:
This is an automous equation,since the right side is a function noly depends on y.And y1 =
0,y2 = 1,y3 = 3 are the critical points in which y1 = 0,y3 = 3 are unstable and y2 = 1 is
asymptotically stable.
6
Let RHS=f (y).
y 00 = f 0 (y)f (y)
f 0 (y) = 3y 2 − 8y + 3
The roots are y4 =
√
4− 7
,y4
3
=
√
4+ 7
.
3
Then in (−∞, y1 ),it’s concave;in (y1 , y4 ),it’s convex;in (y4 , y2 ),it’s concave;in (y2 , y5 ), it’s convex;
in (y5 , y3 ),it’s concave;in (y3 , +∞),it’s convex.
6.Consider the initial value problem:
(
dy
dt
= y 1/2 t ≥ 0, y ≥ 0
y(0) = 0
(a)Show that this problem has more than one solution.
(b)Explain why the fact in (a) does not contradict the well-posedness theorem.
Answer:
(a)y = 0 is obviously the solution.On the other side:
dy/y 1/2 = dt, ⇒ 2y 1/2 = t, ⇒ y = 14 t2 is also a solution.
(b)f (t, y) = y 1/2 is continuous at (0, 0),so the solution exist;∂f /∂y = 21 y −1/2 ,it’s not continuous
at (0, 0) ,thus the solution is not unique.
7.Consider the following initial value problem:
(
u00 + (2π)2 u = cos(ωt)
u(0) = 0,
u0 (0) = 0
with ω being a nonnegative costant.
(a)Determine the value of ω for which the solution remains bounded for all t > 0,and solve the
initial value problem for such ω.
(b)Determine the value of ω for which resonance occurs,then solve the initial value problem for
such ω.
Answer:
7
(a) The general solution to the homogeneous equation is
c1 , c2 ∈ R
u(t) = c1 cos 2πt + c2 sin 2πt,
If ω 6= 2π,then a particular solution has the form
up = A cos(ωt) + B sin(ωt)
It will remain bounded for all t.By substituting it into the equation:A =
Then the general solution to the nonhomegeneous one is :
1
cos(ωt)
u = c1 cos 2πt + c2 sin 2πt + 2
4π − ω 2
Along with the initial condition,the solution is:
1
1
cos
2πt
+
cos ωt
u=− 2
4π − ω 2
4π 2 − ω 2
(b)For ω = 2π,then a particular solution has the form:
up = At cos 2πt + Bt sin 2πt
Substitute it into the equation:A = 0, B =
1
.
4π
The general solution is
u = c1 cos 2πt + c2 sin 2πt +
By initial condition,the solution is:
u=
1
t sin 2πt
4π
1
t sin 2πt
2π
1
,B
4π 2 −ω 2
=0