MATH 316, DIFFERENTIAL EQUATIONS, WINTER 2000

Answers to the Second Midterm

Problem 1 (12 points)

(a) Find all (real or complex) solutions of the form y = e^{rt} which satisfy the following equation.

y′′ − 6y′ + 10y = 0.

Answer: Plug y = e^{rt} into the equation, we get (r^{2} − 6r + 10)e^{rt} = 0. Thus, e^{rt} will be a solution if and only if r^{2} − 6r + 10 = 0. The quadratic formula gives us two answers for r, namely r = 3 + i and r = 3 − i. Thus, the desired solutions are y_{1} = e^{(3+i)t} and y_{2} = e^{(3−i)t}.

(b) Find two linearly independent real solutions of the equation. Answer: Let us write the two solutions from part (a) in terms of their real and imaginary parts:

y_{1} = e^{3t}(cos t + i sin t),
y_{2} = e^{3t}(cos t − i sin t).

Since the system of equations is linear, any linear combination of solutions is again a solution. In particular, by adding y_{1} + y_{2} (and dividing the result by 2), we get a real solution, y_{3} = e^{3t} cos t. By subtracting y_{1} − y_{2} (and dividing the result by 2i), we get another real solution, y_{4} = e^{3t} sin t. These are two linearly independent real solutions.

Problem 2 (12 points)

(a) Carefully sketch the phase plane for the system of equations

\frac{dx}{dt} = −8x + 5y, \frac{dy}{dt} = −10x + 7y.

Indicate clearly in which direction t increases, and make sure that you mark any asymptotes. From your sketch, what can you say about the short-term behavior of the solution with initial condition x(0) = 5, y(0) = 11? What can you say about its long-term behavior? Answer: We write the equations in matrix form:

\frac{d}{dt} (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}−8 & 5 \\ −10 & 7 \end{matrix}) (\begin{matrix}x \\ y \end{matrix}) .

To prepare the sketch, we find the eigenvalues and eigenvectors. The eigenvalues are given by (−8 − λ)(7 − λ) + 50 = λ^{2} + λ − 6 = 0, with solutions λ_{1} = 2 and λ_{2} = −3. The corresponding eigenvectors are v_{1} = (1, 2) and v_{2} = (1, 1). Since the eigenvalues have opposite signs, we have the following situation for the phase plane:

The solution with initial condition x(0) = 5, y(0) = 11 is highlighted. In the short term, x and y increase. In the long term, they both increase towards infinity, asymptotically along the line y = 2x.

(a) Find the general solution to the above system of differential equations.

Answer: Since we have already found a basis of eigenvectors, we can simply write out the general solution, where c_{1} and c_{2} are arbitrary constants:

(\begin{matrix}x \\ y \end{matrix}) = c_{1} (\begin{matrix}1 \\ 2 \end{matrix}) e^{2t} + c_{2} (\begin{matrix}1 \\ 1 \end{matrix}) e^{−3t}.

(c) Find the general solution of the non-homogeneous system of equations

\frac{d}{dt} (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}−8 & 5 \\ −10 & 7 \end{matrix}) (\begin{matrix}x \\ y \end{matrix}) + (\begin{matrix}3 \\ 3 \end{matrix}) .

Answer: We first need to find a particular solution. We try for a constant solution: Setting x′ = y′ = 0, we get

(\begin{matrix}−8 & 5 \\ −10 & 7 \end{matrix}) (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}−3 \\ −3 \end{matrix}) .

Using Gaussian elimination, we find that (x, y) = (1, 1) is a solution. Thus, the general solution to the non-homogeneous system is

(\begin{matrix}x \\ y \end{matrix}) = c_{1} (\begin{matrix}1 \\ 2 \end{matrix}) e^{2t} + c_{2} (\begin{matrix}1 \\ 1 \end{matrix}) e^{−3t} + (\begin{matrix}1 \\ 1 \end{matrix}) .

Problem 3 (12 points) Consider the system of linear differential equations

\frac{dx}{dt} = y \frac{dy}{dt} = \frac{x − ty}{1 − t} .

(a) Find a solution (x(t), y(t)) for which y(t) = 1 is constant.

Answer: If y = 1 is constant, then y′ = 0, and thus x − ty = 0 from the second equation. Hence x = ty = t will satisfy the second equation. We need to check that the first equation is also satisfied; but this is the case since x′ = 1. Thus (x, y) = (t, 1) is a solution.

(b) Verify that (x, y) = (e^{t}, e^{t}) is another solution.

Answer: We need to check that each equation is satisfied. For the first equation, left and right-hand sides are equal to e^{t}. For the second equation, the left-hand side is e^{t}, while the right-hand side is (e^{t} − te^{t})/(1 − t) = e^{t}(1 − t)/(1 − t) = e^{t} (unless t = 1, in which case the equation is undefined).

(c) Find a solution which satisfies the initial conditions x(0) = 1 and y(0) = 0.

Answer: Since we are dealing with a linear system here, linear combinations of solutions are again solutions. Thus, the general solution is (x, y) = c_{1}(t, 1) + c_{2}(e^{t}, e^{t}), where c_{1} and c_{2} are arbitrary constants. We plug in the initial condition t = 0, x = 1, y = 0 to determine c_{1} and c_{2}: (1, 0) = c_{1}(0, 1) + c_{2}(1, 1). This system is easily solved, and we find that c_{2} = 1 and c_{1} = −1. Thus, the desired solution is (x, y) = −1 · (t, 1) + 1 · (e^{t}, e^{t}) = (e^{t} − t, e^{t} − 1).