MATH2270

FW00

Solution to Assignment 3

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

(i) xy´ + 3y = 2x3/2 + x2;

Writing this as y´ + 3y/x = 2x1/2 + x we have a linear equation in standard form with

so that the integrating factor is x3 and the solution is

or y = 4/9 x3/2 + x2/5 + C/x3.

(ii) y´ - 2cot(x) y = sin2(x)

This linear equation has

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

giving y = (x + c) sin2(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 + sin2(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) x2 y´+ 2xy - y3/2 = 0;

Writing this as y´ + (2/x) y = y3/2 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 = 16x2(1+Cx2)-2

(ii) y´ + 2xy + xy4 = 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(-3x2). The solution is

or u = -1/2 + c exp(3x2).

Thus y = u-1/3 = [-1/2 + c exp(3x2)]-1/3

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

(i) (2x3 + 3y) dx + (3x + y - 1) dy = 0;

Let M = 2x3 + 3y and N = 3x + y - 1. ThusSince these are equal the DE is exact and

.

Now

Thus

Thus the implicit solution is x4/2 + 3xy + y2/2 - 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 e2t 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 - y2).

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

The equilibrium solutions are the roots of g(y) = y(4 - y2) = 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-3y´ - 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 e8t 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) d2y/dx2 = 2x2 dy/dx - 4xy3;

Let z = dy/dx. Then we get the system dy/dx = z , dz/dx = 2x2z - 4xy2 of two first order DEs.

(ii) 2x d2y/dx2 - 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/(y2 - x).

The system is dx/dt = y2 - 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.