Solutions toTerm Test I
test was marked out of 60.
1. For each of the following differential equations find the analytic family of solutions which satisfies the differential equation and indicate any symmetries which the diferential equation possesses. Also find the explicit solution which satisfies the initial conditions given and any equilibrium solutions:
(a) (7 marks) y(1) = 3
Analytic family of solutions:
Applying initial condition yields c = -7/6 which gives the explicit solution y = -3 + 6/(7-6x).
are no symmetries. y = -3 is an equilibrium solution.
(b) (8 marks) y(1) = 3
Analytic family of solutions:
Applying initial condition yields C = -e-2 which gives the explicit solution y = 4 - xe2x-2.
are no symmetries. y = 4 is an equilibrium solution.
(c) (11 marks) y(1) = 1
this as we
see this is a DE with homogeneous coefficients. Letting y = zx then y´
= xz´ + z and the DE becomes xz´ = z2.
The analytic family of solutions is:
Applying initial condition yields c = -1 which gives the explicit solution y = x/(1-ln|x|).
The DE is symetric about the origin. y = 0 is an equilibrium solution.
DE could also be solved as a Bernoulli equation.
(d) (8 marks) y(0) = 1
Since e(x+y) = ex ey we have the analytic family of solutions:
Applying initial condition yields c = -1 - e-1 which gives the explicit solution y = -ln(1 + e-1 - ex).
are no symmetries. There are no equilibrium solutions.
2. (a) (3 marks) Use the Existence and Uniqueness Theorem to determine where the differential equation
has unique solutions.
The rhs g(y) = (y+1)1/3 is defined for all y.which has a singularity at y = -1. The E&U Thm is
(b) (8 marks) Find the explicit solution to this differential equation for all x that satisfies the initial condition y(2) = 7.
Solving the DE gives
Noting the lhs is always positive, the solution is valid only for x > -c.
Applying the initial condition gives c = 6 - 2 = 4. We also note that y = -1 is an equilibrium solution.
explicit solution for all x is y = -1 for x < -4 and y = -1 + [(2x+8)/3]3/2
for x > -4.
(8 marks) A tank holds 500 litres of water. At time t = 0 water is poured
into the top of the tank at the rate of 2 litres per second and a drain
is opened at the bottom of the tank. Water flows out of the drain at the
rate of 0.5% of the volume of water in the tank per second. Write down
the differential equation for the amount of water in the tank at time t
> 0 and find the explicit solution that satisfies the initial condition.
How much water will be in the tank after a long time?
This is a simple rate problem where the rate of change of the volume of water, v, in litres equals inflow minus outflow, i.e.
or v = 400 + Ce-0.005t. Applying the initial condition that v(0) = 500 we find C = 100. Thus the volume of water in the tank at time t is 400 + 100e-0.005t.
After a long time the exponential term goes to zero leaving 400 litres in the tank. This is also the equilibrium solution of the DE.
we could also solve the DE as a linear DE.
4. (7 marks) Find the orthogonal trajectories of the family of curves y = p sin(x) where p is a parameter taking on different values.
y´ = p cos(x) but p = y/sin(x) so that y´ = y cot(x)
For the orthogonal trajectories we replace y´ by -1/y´ to get y´ = -tan(x)/y. Solving this DE yields
5. (4 marks) Indicate whether the following statements about the differential equation y´ = g(y) are true or false:
(i) The equilibrium solutions are always isoclines. true
(ii) The isoclines are always equilibrium solutions. false
(iii) If g(y) an even function of y, the family of solutions of the differential
equation is symmetric about the x-axis. false
An explicit solution of the differential equation can always be found.
6. (6 marks) The following plot contains several members of the family of solutions of the differential equation:
(a) Indicate on the plot the regions in the x-y plane where the solutions are monotonically increasing. Also indicate regions where the solutions are concave down.
(b) Indicate which curve corresponds to the solution with initial condition y(0) = 2.
Draw the slope field at the point (1, -3).
plot posted on notice board near 122 Petrie.