**MATH2270**

**FW00**

**Solution
to Assignment 3 **

1. (a) Find the family of explicit solutions for the following linear differential equations:

(i) xy´ + 3y = 2x^{3/2} + x^{2};

Writing this as y´ + 3y/x = 2x^{1/2} + x
we have a linear equation in standard form with

so that the integrating factor is x^{3} and the
solution is

or y = 4/9 x^{3/2} + x^{2}/5 + C/x^{3}.

(ii) y´ - 2cot(x) y = sin^{2}(x)

This linear equation has

so
the integrating factor is sin^{-2}(x) and the solution is

giving y = (x + c) sin^{2}(x)

(b) Apply the Existence and Uniqueness Theorem to the differential equation in part (ii). Investigate the nature of the family of solutions at points where the theorem is not satisfied.

Since y´ = 2cot(x) y + sin^{2}(x) = g(x,
y) and

the E&U Thm is satisfied everywhere except when cot(x) is singular, i.e. when sin(x) = 0. But when sin(x) = 0, y is always zero for any value of c. Thus the solutions are not unique at these points.

2. Find the family of explicit solutions for the following Bernoulli differential equations:

(i) x^{2} y´+ 2xy - y^{3/2} = 0;

Writing this as y´ + (2/x) y = y^{3/2} or
y^{-3/2} y´ + (2/x) y^{-1/2} = x^{-2}
we see this is a Bernoulli equation. Let

u = y^{-1/2} so that u´ = -(1/2) y^{-3/2}
y´ and the DE becomes -2u´ + 2u/x = x^{-2} or u´
- (1/x) u = -(1/2) x^{-2}

which is a linear DE with so that the integrating factor is 1/x and we have the solution or u = 1/(4x) + cx.

But y = u^{-2} = 16x^{2}(1+Cx^{2})^{-2}

(ii) y´ + 2xy + xy^{4} = 0;

This DE becomes y^{-4 }y´ + 2x y^{-3}
= -x. Let u = y^{-3} so that u´ = -3y^{-4} y´
and we get u´ - 6xu = 3x. Thus

and
the integrating factor is exp(-3x^{2}). The solution is

or
u = -1/2 + c exp(3x^{2}).

Thus y = u^{-1/3} = [-1/2 + c exp(3x^{2})]^{-1/3}

3. Show that the following differential equations are exact and find the family of implicit solutions:

(i) (2x^{3} + 3y) dx + (3x + y - 1) dy = 0;

Let M = 2x^{3} + 3y and N = 3x + y - 1.
ThusSince
these are equal the DE is exact and

Now

Thus

Thus the implicit solution is x^{4}/2 + 3xy +
y^{2}/2 - y = C

Alternately we can find H as Now

Thusgiving the same solution as above.

(ii) (cos(y) + y cos(x)) dx + (sin(x) - x sin(y)) dy = 0;

Let M = cos(y) + y cos(x) and N = sin(x) - x sin(y).

Then

Since these are equal, the DE is exact and

Now

and the implicit solution is x cos(y) + y sin(x) = C.

Alternately, and

giving the same answer as above.

4.
An Electrical RL circuit has an inductance L = 3 henrys and a
resistance R = 6 ohms. If the applied voltage E = sin(t) e^{-2t}
volts find the current I (in amperes) in the circuit if I(0) = 0.

The
DE for this circuit is L I´ + R I = E or
.
This is a linear DE with integrating factor e^{2t} so that
the solution is
or

I
= -(1/3)e^{-2t} cos(t) + c e^{-2t}.

Applying the initial condition I(0) = 0 yields c = 1/3 so that
I = (1/3)[1 - cos(t)] e^{-2t}.

5. Investigate the logistic equation where the
overcrowding term varies as the cube of the population. Consider the
particular differential equation y´ = y(4 - y^{2}).

(a) What are the equilibrium solutions? Classify them as stable or unstable.

The equilibrium solutions are the roots of g(y) = y(4 -
y^{2}) = 0, i.e. y = 2, 0, -2. Since

which is negative for y = 2 or -2 and positive for y = 0. Thus the equilibrium solutins y = 2 and y = -2 are stable and y = 0 is unstable.

(b) Solve this differential equation as a Bernoulli equation. Show that your solutions have the behaviour predicted in part (a).

The DE becomes y^{-3}y´ - 4y^{-2}
= -1. Let u = y^{-2}. Then u´ = -2y^{-3} y´
and the DE becomes u´ + 8u = 2. This is a linear DE with
integrating factor e^{8t} so the solution is
or

u = 1/4 + c e^{-8t}.
Thus y = u^{-1/2}
=
.

As t becomes large the exponential term becomes very small so that the y approaches one of the stable equilibrium solutions y = 2 or -2 as expected.

6. (a) Convert the following second-order differential equations to a system of first-order differential equations:

(i) d^{2}y/dx^{2} = 2x^{2 }dy/dx
- 4xy^{3};

Let z = dy/dx. Then we get the system dy/dx = z , dz/dx
= 2x^{2}z - 4xy^{2} of two first order DEs.

(ii) 2x d^{2}y/dx^{2} - 2dy/dx + 4xy =
sin(2x), y(0) = 1, y´(0) = -2.

Let z = dy/dx. Then we get the system dy/dx = z , dz/dx = z/x - 2y + sin(2x)/2x.

The initial conditions become y(0) = 1, z(0) = -2.

(b) Convert the following first-order differential equations into a system of autonomous first-order differential equations where both x and y depend on the independent variable t;

(i) dy/dx = (x + y)/(x - y);

The system is dx/dt = x - y, dy/dt = x + y

(ii)
dy/dx = 1/(y^{2} - x).

The system is dx/dt = y^{2} - x, dy/dt = 1

7. (a) Use MAPLE to plot solutions to the following first-order systems. Use both scene = [x,y] and scene = [t,x] to produce two plots showing different aspects of the behaviour of these systems.

(i) dx/dt = sin(x) + cos(y); dy/dt = cos(x) - sin(y);

(ii) The system obtained in question 6 (a) (i);

(iii) The system obtained inquestion 6 (b) (ii).

See MAPLE output

(b) On the orbit plot of part (a) (iii) indicate the
**solution of the original differential equation **which satisfies
the initial condtion y = 0 at x = 1. You will have to include this
initial condition on your plot command. How does this plot differ
from the MAPLE plot you obtained in question 1 (d) of Assignment 2?

See MAPLE output

(c) Use the Existence and Uniqueness Theorem for systems to show that the system in part (iii) has a unique solution everywhere. In your plot of x vs t in part (a) (iii) include the initial conditions y = 1 when x = 0 and y = -1 when x = 0. Why do these curves cross in the plot if the solutions are unique?

Since the right-hand sides of the system in 6 (b) (ii) have well behaved partial derivatives, the E&U Thm is satisfied everywhere. Thus the solution curves do not cross. But these curves are three-dimensional (x, y, t) and we are looking at their projection in the x-t plane where they appear to cross.