MATH2270

FW00

Solutions to Assignment 2

1. (a) Use the Existence and Uniqueness Theorem to determine where the following differential equations have unique solutions:

i) ii)

(i) If we let so the E&U Thm is satisfied for y < 2. There is no solution for y > 2.

(ii) The E&U Thm is satisfied everywhere except on the curve y2 = x.

(b) Find the regions of the x-y plane where the solutions to the differential equations are monotonically increasing or decreasing. Also find any equilibrium solutions.

(i) the solution is monotonically increasing for y < 2. y = 2 is an equilibrium solution.

(ii) the solution is increasing for y2 > x and decreasing for y2 < x. There are no equilibrium solutions.

(c) Find a solution to the differential equation (i) with the initial condition y(0) = 1 that is valid for all x. Be careful to check that you have the correct solution.

The solution is or -4(2-y)1/4 = x + c. Note that since the lhs is negative we must have x < -c.

Solving for y yields y = 2 - (x+c)4/28. Applying the initial condition yields c = -4 which means the solution is y = 2 - (x-4)4/28 for x < 4.

Note that when x = 4, y = 2 which is the equilibrium solution. Thus the solution for x > 4 is y=2.

(d) Use MAPLE to plot solutions of the differential equation (ii) with initial conditions of the form y(0) = y0. Chose at least one value of y0 in each of the following intervals: 1 < y0; -1 < y0 < 1; y0 < -1. Do you think the curves that MAPLE produces are correct? Explain what is going on.

See MAPLE output.

2. (a) Find the regions of the x-y plane where the solutions to the following differential equation are monotonically increasing or decreasing. Derive any symmetries that these differential equations possess and find any equilibrium solutions.

i) ii)

(i) The solutions are increasing for |y| > |x| (i.e for y > x AND y > - x OR y < x AND y < -x). It is decreasing for |y| < |x| (y < x AND y > -x OR y > x AND y < -x).

The DE is symmetric about the origin.

There are no equilibrium solutions.

(ii) The solutions are increaing for 2 sin(y) + x2y > 0 and decreasing for 2 sin(y) + x2y < 0. (We cannot solve this equation for y in terms of x but we could solve for x in terms of y. Since |sin(y)/y| < 1 then the slope is always positive for x2 > 2 and y > 0 and always negative for x2 > 2 and y < 0. It is positive for 0 < y <Pi, 2Pi < y < 3Pi, etc. It is negative for -Pi < y < 0, -3Pi < y < -2Pi, etc.)

The DE is symmetric about the x-axis.

y = 0 is an equilibrium solution.

(b) Use MAPLE to plot several solution curves for each of these differential equations. Choose initial values to illustrate the properties derived in part (a).

See MAPLE output.

(c) Use the comparison theorem to show that the solution of the differential equation given in (i) with initial condition y(0) = 2 has a vertical asymptote in the interval 0.5 < x < 0.55 by noting that for

0 < x < 1, y2 - 1 < y2 - x2 < y2.

Let y1 be the solution of the DE y´ = y2 - 1 and y2 be the solution of the DE y´ = y2. Thus by the Comparison Thm the solution to (i) lies between these two solutions. Since

we have y1 = (1+Ce2x)/(1-Ce2x). Applying the initial condition yields C = 1/3 so that y1 has a vertical asymptote when e2x = 3 or x = ln(3)/2 = ~ 0.55.

From the lectures we know that y2 = -1/(x+c) and c = -1/2 to satisfy the initial condition. Thus y2 has a vertical asymptote at x = 0.5.

Thus the solution to (i) must have a vertical asymptote for 0.5 < x < 0.55.

3. (a) Find the analytic family of solutions to the following separable differential equation:

What symmetries does this differential equation have? What are the equilibrium solutions? Use MAPLE to plot several members of this family that illustrate these properties.

The family of solutions iswhich is an implicit solution (though we could solve for x in terms of y).

The DE is symmetric about the x-axis, the y-axis and the origin.

y = 0 is an equilibrium solution.

See MAPLE output for thje plot.

(b) A chemical reaction depends on the amount of the reagent Y and the temperature of the reagent. Initially there is 10 kg of the reagent in solution and the temperature is 300 K. If the reagent decreases at a rate of 0.1% of the reagent present times the temperature of the solution per minute and the solution is heated at the rate of 2 K per minute, find the amount of reagent in the solution as a function of time.

Let y be the number of kilograms of reagent in the tank at time t and let T be the temperature in kelvins at time t. Then the rate of change of the reagent is -0.001yT. The temperature is given by T = 300 + 2t (this can be derived from the DE dT/dt = 2 with T(0) = 300). Thus the DE for the reagent is

which is a separable DE with solution

Thus y = Cexp(-0.3t-0.001t2) where C = 10 from the initial conditions.

4. Find analytic families of solutions for the following differential equations with homogeneous coefficients. Also find all solutions of the form y = mx for these differential equations where m is a constant.

(a)

(b)

In both these cases we let y = xz so that y´=xz´+z

(a) This DE becomes xz´+z = z - 1/z or z´ = -1/zx. This is separable with solution

Substituting for z we get y2 = x2(C - ln x2). This gives two families of solutions

Letting y = mx yields the equation m2 = m2 - 1 which has no solutions.

(b) This DE becomes xz´+z = z + z ln z. This is separable with solution

or z = exp(Cex). Substituting for z gives y = x exp(Cex).

Letting y = mx yields the equation m = m + m ln(m) or m ln(m) = 0 which has solutions m = 0 or m = 1, i.e. the DE has solutions y = 0 and y = x. The first is the equilibrium solution and the second comes from the family of solutions with C = 0.

5. (a) Find the orthogonal trajectories of the family of curves given by the following equations where p is a paramter which takes on different values:

i) y = pe-x ii) x2 + py2 = 1

(i) y´ = -pe-x = -y. Thus the orthogonal trajectories satisfy the DE y´ = 1/y or

(ii) 2x + 2pyy´ = 0, i.e. y´ = -x/py = -xy/(1-x2). Thus the orthogonal trajectories satisfy the DE

y´ = (1-x2)/xy which is separable with solution

or

(b) See MAPLE output.