Section Check In –
Core Pure Differential Equations
Questions
1.*
(i) The resistance to motion for a person falling through the air with a parachute is modelled
as being proportional to the square of the velocity, v. Show that the motion of a particular
person falling through the air with a parachute may be modelled by the differential
dv
equation v  g  kv 2 where x is the displacement of the person, g is the acceleration
dx
due to gravity and k is a constant.
(ii) Find the general solution of the differential equation as an expression for x in terms of v.
2.*
Choose an appropriate method to find the general solution to
3.*
(i) Write down the auxiliary equation for
3
dy 2
 y  e x .
dx x
d2 y
dy
 5  3 y  0 and hence find the
2
dx
dx
general solution.
(ii) We require that y  0 as x  and y  5 when x  0 . Using this information and your
answer to i), find the particular solution.
4.*
(i) Find the general solution to
d2 y
 4y  0 .
dt 2
(ii) Find the particular integral of
d2 y
 4 y  4t 2  2 and hence write down the full general
2
dt
solution.
5.*
Find the solution to
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d2 y
 4 y  4t 2  2 given that y  1 when t  0 and y  2 when t  1 .
2
dt
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6.*
The equation 
d2 y
dy
     0 is studied, where  ,  and  are constants. Make a sketch
2
dx
dx
of the solutions in the following cases:
(i)  2  4 0 ,  ,   0
(ii)  2  4 0 ,  0 ,  0
7.*
The equation for damped Simple Harmonic Motion of a bridge is given by
d2 x
dx
 k 2x  0
2
dt
dt
(i) Explain the physical significance of the constants k and  2 .
(ii) Write down the conditions on k and  for under-damping.
8.*
Although not part of the syllabus. This second-order ODE with simple polynomial coefficients
can be solved simply using the method taught in this module and provides a nice test of
understanding in verifying solution to differential equations.
By direct substitution, verify that y  x a for constant  , is a solution to
x2
9.*
d2 y
dy
 4 x  4 y  0 . State the possible values of  .
2
dx
dx
A model is proposed to investigate the population of foxes x and chickens y in the form of a
system of differential equations:
x  ax  by
y  cx  dy
for constants a , b c, d . Initially the population of foxes is x0 and chickens is y0 .
(i) Solve the system of equations for a  2, b  4, c  1, d  1 in terms of x0 and y0 .
(ii) Describe the long-term behaviour of the fox and chicken populations.
(iii) Give a possible criticism of the suitability of this model.
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10.* An object is dropped into the sea from an oil rig platform. When it hits the surface, it is
travelling at a speed of 10 ms-1. The time is measured from when the object hits the surface
at a depth x  0 . The resistance force FR is modelled by FR 
dx
where x is the distance
dt
travelled from the oil rig platform. The only other force acting on the object as it falls is
gravity. The resistance force is thought to be double the speed of the object and the object
has a mass of 10kg. (take g  10 ms-2).
(i) Write down an equation of motion using Newton’s second law.
(ii) Solve this equation using the initial conditions.
Extension
Investigate the solutions to
dx
 ax  by
dt
dy
 cx  dy
dt
for different a, b, c, d .
Can you find conditions on a, b, c, d so that (provided the arbitrary constants are chosen carefully)
(i)
x and y  0 as t  
(ii)
x and y   as t  
(iii)
x and y are oscillatory as t  
Worked solutions
(i) Using Newton’s second law, ma  mg  Kv 2 where m is the mass of the person and
K is a constant
1.
For a particular person, m is a constant so dividing through by m gives
dv
a  g  kv 2 so v  g  kv 2
dx
(ii) Separating variables
dv
 v g  kv   dx
2
C
2.
1
ln g  kv2  x where C is a constant.
2k
This is solved using the method of integrating factor. Comparing to the general form of a
linear ODE:
dy
 P  x y  Q  x
dx
p x dx
2 x 1dx
 e2 ln x  x 2
The integrating factor is R  x   e    e 
3
dy
 2 xy  x2e x
dx
3
d 2
The LHS can be written as a perfect derivative:
x y   x 2 e x

dx
3
1 3
Integrating with respect to x : x 2 y   x 2 e x dx   e x  C , where C is a constant of
3
Multiplying the ODE by R  x   x 2 : x2
integration.
Finally, we have the general solution: y  
3.
i)
1
1  3 x3 C
e
 2
3x 2
x
The auxiliary equation is found by substituting y  e x into the ODE:
 2e x  5 e x  3e x  0
The exponentials cancel and we are left with the quadratic:  2  5  3  0
5  25  12 5  37

2
2
Therefore, the general solution is y  Ae1x  Be2 x
5  37
5  37
Where 1 
and 2 
. [the choice of which is which is arbitrary]
2
2
This has roots  
ii)
It is easy to see that 1  0 and 2  0 . Therefore, for solutions to tend to zero as y   ,
we must have the coefficient of e1x be zero. This means A  0 and hence y  Be2 x
using the initial conditions: 5  Be0 and therefore, the particular solution is y  5e2 x .
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4.
i)
The auxiliary equation is  2  4  0 which has roots at   2i and hence the general
solution is y  A cos 2t  B sin 2t .
ii)
The particular integral is found by substituting y  Ct 2  Dt  E into the ODE:
y  Ct 2  Dt  E
y  2Ct  D
y  2C


and hence substituting: 2C  4 y  Ct 2  Dt  E  4t 2  2
Rearranging the LHS: 4Ct  4 Dt  2  C  2E   4t 2  2
2
and hence equating coefficients: 4C  4 , D  0 , 2 C  2E   2
which are solved by C  1, D  0 and E  0 and hence the general solution is:
y  C.F  P.I
y  A cos2t  B sin 2t  t 2
5.
The auxiliary equation is  2  4  0 which has roots at   2 and hence the general
solution is y  Ae2t  Be2t
The particular integral is found by substituting y  Ct 2  Dt  E into the ODE:
y  Ct 2  Dt  E
y  2Ct  D
y  2C


and hence substituting: 2C  4 y  Ct 2  Dt  E  4t 2  2
Rearranging the LHS: 4Ct 2  4 Dt  2  C  2 E   4t 2  2
and hence equating coefficients: 4C  4 , D  0 , 2 C  2E   2
which are solved by C  1, D  0 and E  1 and hence the general solution is:
y  C.F  P.I
y  Ae2t  Be2t  t 2  1
y  1 when t  0
1  A  B  1  A  B  0 so B   A
y  2 when t  1
2  Ae 2  Be 2  2  Ae 2  Be 2  0


Ae 2  Ae 2  0  A e 2  e 2  0
e 2  e 2  0  A  0 so B  0
y  t 2  1
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6.
The auxiliary equation is  2      0 which has solution  
i)
   2  4
.
2
For this case, there are 2 distinct real roots of the auxiliary equation. Additionally, we
know that because 𝛼 and 𝛾 are both positive then the square root term in the formula
has a magnitude smaller than  , i.e.  2  4  
and hence the two real roots are both negative. Therefore, the graph will look like:
ii)
In this case, the roots are complex. The real part of the roots is given by:
    

0
2
Because   0 and   0 . Therefore, the solution will oscillate in a decaying manner
around y  0 as x  :
7.
i)
k is the damping coefficient – a measure of the bridge’s resistance to motion via airresistance/dashpot.
The constant
ii)
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
2
is the natural frequency of the oscillations of the un-damped system.
For under-damping the solution is oscillatory. Hence the auxiliary equation:
 2  k    2  0 must have complex roots and hence k 2  4 2 0 .
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8.
To verify a solution first we differentiate the possible form of the solution:
y  x
y   x 1
y    a  1 x  2
Substituting these expressions into the ODE gives   a 1 x 2 x 2  4 x 1 x  4 x  0
which can be simplified to   a 1  4  4  x  0 .
We have therefore that the term in brackets has to be zero. Hence   a  1  4  4  0
Simplifying:  2  3  4  0 which has two solutions   1, 4
and hence the general solution is y  Ax1  Bx4
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9.
i)
The system becomes:
The first equation is rearranged for y :
It is then differentiated:
These two expressions are substituted
into the second equation to get:
which is simplified to
2    6  0
This has auxiliary equation:
which has roots   2 and   3
and hence the complementary function is
x  Ae2t  Be3t
and hence using the expression for y above we have
5
y   Be3t .
4
which is simplified to
Now using the initial conditions we have that at t  0 , y  y0 , and hence B  
Using the solution and initial conditions for x we have
4
x0  A  y0
5
and therefore
4
A  x0  y0
5
4 
4

x   x0  y0  e2t  y0 e3t , y  y0e3t .
5
5


and the general solution is
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y
5 0
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ii)
The chickens for any y0  0 will die out as the solution is decaying exponential. The
foxes population will increase provided that x0 
4
y  0.
5 0
If this is not satisfied then the foxes will also die out as there will only be a decaying
exponential.
iii) If the initial chicken population is y0  0 then the fox numbers will evolve like x  x0e2t
i.e. they will experience exponential growth; unrealistic in the long term.
10.
i)
Using Newton’s second law we have F  FGrav  FR  ma
The constant of proportionality for the air resistance force is 2 and the mass is 10 kg
hence
ma  mg  2
10
ii)
dx
dt
d2 x
dx d 2 x 1 dx

10

10

2
 2
 10
dt
dt 2
dt 5 dt
We solve the homogenous equation first.
This could be done by using an auxiliary equation or, alternatively, as follows.
We write this as a first-order ODE:
where v 
dv
1
 v
dt
5
dx
is the speed. Therefore, separating the variables:
dt
1
5
Evaluating the integrals: ln v   t  C and hence v  Ae
1
1
 t
5
Using the initial conditions we have that 10  Ae and hence v
0
1
 vdv   5dt
1
 t
 10e 5
1
1
 t
 t
dx
 10e 5 is solved directly by integration to get x  2e 5  C
dt
Using the initial conditions that at t  0 , x  0 we have 0  2  C and hence C  2 .
Now the equation:

The homogenous solution is therefore x  2 1  e


1
 t
5

.


The particular integral is found by substitution of x  At as the RHS of the ODE has a
similar form to one of the homogenous solutions; ie. A constant. Substituting x  At
1
5
into the ODE gives 0  A  10 and hence A  50 and the full solution is

x  2 1  e


1
 t
5

  50t


Extension
ii)
a  d  0, (a  d )2  4(ad  bc)  a  d
ad 0
iii)
(a  d )2  4(ad  bc)
i)
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