18.034 MIDTERM 2: SKETCHES OF SOLUTIONS

Explain your answers clearly; show all steps. Calculators may not be used. All problems have equal value. Please put your name on every sheet. Good luck!

1. (a) y_{1}, y_{2}, and y_{3} are 3 solutions of the differential equation (1 − t)y′′′ + y′′ + t^{2}y′ + t^{3}y = 0 on the interval 1 < t < ∞. Calculate the function W (y_{1}, y_{2}, y_{3})(t) given that W (y_{1}, y_{2}, y_{3})(2) = 3.

Solution. Rewrite the differential equation as y′′′ + \frac{1}{1−t}y′′ + \frac{t^{2}}{1−t}y′ + \frac{t^{3}}{1−t}y = 0. Then by Abel’s theorem, W = c exp(− ∫ \frac{1}{1−t}dt) = c(1 − t) for some constant t. From the condition W (2) = 3, we get W (t) = 3(t − 1).

(b) The equation y′ + a(x)y = 0 has for a solution

ϕ(x) = e^{− ∫_{x_{0}}^{x} a(t)dt}.

(Here let a be continuous on an interval I containing x_{0}.) This suggests trying to find a solution of

L(y) = y′′ + a_{1}(x)y′ + a_{2}(x)y = 0

of the form

ϕ(x) = e^{∫_{x_{0}}^{x} p(t)dt}

where p is a function to be determined. Show that ϕ is a solution of L(y) = 0 if and only if p satisfies the first-order non-linear equation y′ = −y^{2} − a_{1}(x)y − a_{2}(x). (Remark: This last equation is called a Riccati equation.)

Solution. ϕ(x) = e^{∫ p(t)dt}, so ϕ′(x) = p(x)e^{∫ p(t)dt} and ϕ′′(x) = p′(x)e^{∫ p(t)dt} + p(x)^{2}e^{∫ p(t)dt}. If ϕ satisfies the differential equation, then

e^{∫ p} (p′ + p^{2} + a_{1}p + a_{2}) = 0

from which the result follows.

2. (a) Consider the equation y′′ − \frac{2}{x^{2}} y = 0 (for 0 < x < ∞). Find all solutions. (Hint: Try functions of the form y = x^{r}. How do you know you’ve found all the solutions?)

(b) Find all solutions to the equation y′′ − \frac{2}{x^{2}} y = x. Hint: Use “variation of parameters”. Suppose ϕ_{1} and ϕ_{2} are linearly independent solutions to the homogeneous version of the equation (see (a)). Look for a solution of the form ϕ(x) = u_{1}(x)ϕ_{1}(x) + u_{2}(x)ϕ_{2}(x).

Date: April 14, 2000, 1:05-1:55 pm.

Solution. (a) x^{2} and 1/x both work, so C_{1}x^{2} + C_{2}/x work. These are all the solutions by the existence and uniqueness theorem (see for example Theorem 3.2.4).

(b) The general answer is x^{3}/4 + Ax^{3} + B/x. This can be found using the “Variation of Parameters” formula, see Theorem 3.7.1.

Alternatively, here is the argument, explicitly. Let ϕ_{1}(x) = x^{2}, ϕ_{2}(x) = 1/x be a basis for the space of solutions. We seek a single solution to the differential equation, as we already know the solutions to the homogeneous version.

We look for a solution ϕ = u_{1}ϕ_{1} + u_{2}ϕ_{2}, such that

u′_{1}ϕ_{1} + u′_{2}ϕ_{2} = 0 (1)

Then ϕ′ = u_{1}ϕ′_{1} + u_{2}ϕ′_{2}, and

ϕ′′ = (u_{1}ϕ′′_{1} + u_{2}ϕ′′_{2}) + (u′_{1}ϕ′_{1} + u′_{2}ϕ′_{2}).

As ϕ” − (2/x^{2})ϕ = x, we have

u′_{1}ϕ′_{1} + u′_{2}ϕ′_{2} = x. (2)

Rewriting (1) and (2):

u′_{1}x^{2} + u′_{2}/x = 0
u′_{1}(2x) + u′_{2}(−1/x^{2}) = x

and solving this systems gives u′_{1} = \frac{1}{3}, u′_{2} = −\frac{1}{3}x^{2}.

Take u_{1} = \frac{1}{3}x, u_{2} = −\frac{1}{12}x^{4}. Then

ϕ = \frac{1}{3}x(x^{2}) − \frac{1}{12}x^{4}(\frac{1}{x}) = \frac{1}{4}x^{3}.

To be safe, we check that ϕ(x) = x^{3}/4 really does satisfy the differential equation.

3. Iterate x → \sqrt{1 + x}. Start with x = 0. What happens?

Solution. The Contraction Mapping Theorem applies to the interval 0 ≤ x < ∞, as if f(x) = \sqrt{1 + x} then f maps the interval to itself, and f′(x) = 1/(2\sqrt{1 + x}), so |f′(x)| ≤ 1/2. Hence we approach a fixed point x_{0}, satisfying x_{0} = \sqrt{1 + x_{0}}. Squaring and solving, we get x_{0} = (1 ± \sqrt{5})/2. As x_{0} must lie in the interval, x_{0} = (1 + \sqrt{5})/2, the golden mean.

4. (a) State the Existence and Uniqueness Theorem for differential equations of the form y′ = f(x, y).

(b) Consider the differential equation y′ = t^{2}(y + 1) on the interval ℝ, with initial condition y(0) = 0. Find a solution y = ϕ(t) defined for all t ∈ ℝ. If the first few Picard iterates (used in the proof of the Existence and Uniqueness Theorem described in (a)) are ϕ_{0}(t) = 0, ϕ_{1}(t), ϕ_{2}(t), find ϕ_{1}(t) and ϕ_{2}(t).

(c) Explain why the ϕ_{1}(t) and ϕ_{2}(t) you found are approximations to ϕ(t).

Solution. (a) See practice midterm.

(b) From y′/(y + 1) = t we have ln |y + 1| = t^{2}/3, from which y = e^{t^{3}/3} − 1.

ϕ_{k+1}(t) = ∫_{0}^{t} s^{2}(ϕ_{k}(s) + 1)ds, from which inductively ϕ_{1}(t) = t^{3}/3, ϕ_{2}(t) = t^{3}/3 + t^{6}/18.

(c) The power series expansion (or Taylor series expansion) for e^{t^{3}/3} begins

1 + \frac{(t^{3}/3)}{1!} + \frac{(t^{3}/3)^{2}}{2!} + \frac{(t^{3}/3)^{3}}{3!} + ⋯,

so the power series expansion for e^{t^{3}/3} − 1 begins

\frac{(t^{3}/3)}{1!} + \frac{(t^{3}/3)^{2}}{2!} + \frac{(t^{3}/3)^{3}}{3!} + · · · .

In this case, the first few Picard iterates (and indeed all iterates) are partial sums of the power series.

5. Consider the equation y′′ + cos(x)y′ + sin(x)y = 0.

(a) Let ϕ(x) be a nontrivial solution, and let ψ(x) = ϕ(x + 2π). Prove that ψ(x) is also a solution.

(b) Show that ϕ(x) is a periodic solution of period 2π if, and only if, ϕ(0) = ϕ(2π) and ϕ′(0) = ϕ′(2π).

(c) Let ϕ_{1}(x), ϕ_{2}(x) be two solutions satisfying ϕ_{1}(0) = 1, ϕ′_{1}(0) = 0, ϕ_{2}(0) = 0, ϕ′_{2}(0) = 1. Show that there are constants a and b such that

ϕ_{1}(x + 2π) = aϕ_{1}(x) + bϕ_{2}(x).

(Hint: See (a).)

Solution. (a) ψ(x) satisfies the differential equation y′′ + cos(x − 2π)y′ + sin(x − 2π)y = 0, which is the original differential equation.

(b) If ϕ(x) is a periodic solution of period 2π, then by periodicity, ϕ(0) = ϕ(2π) and ϕ′(0) = ϕ′(2π). Conversely, if ϕ(0) = ϕ(2π) and ϕ′(0) = ϕ′(2π), then ϕ(0) = ψ(0) and ϕ′(0) = ψ′(0). As ψ(x) and ϕ(x) have the same initial conditions and satisfy the same differential equation, by the Existence and Uniqueness Theorem (for second-order linear equations with continuous coefficients), ϕ(x) = ψ(x) = ϕ(x + 2π), i.e. ϕ is periodic.

(c) By the Wronskian test, ϕ_{1}(x) and ϕ_{2}(x) are linearly independent solutions of the differential equation, and hence form a basis for the solution space. As ϕ_{1}(x + 2π) is also a solution, it is a linear combination of ϕ_{1} and ϕ_{2}.

6. Let A = ( \begin{matrix}3 & 1 \\ −5 & −3 \end{matrix} ). Find the eigenvalues of A. Find eigenvectors of A

corresponding to each of the eigenvalues. Calculate A^{2000}.

Solution. The eigenvalues are λ_{1} = 2 and λ_{2} = −2, and the corresponding eigenvectors are v⃗_{1} = ( \begin{matrix}1 \\ −1 \end{matrix} ) and v⃗_{2} = ( \begin{matrix}1 \\ −5 \end{matrix} ) respectively. (Any nonzero multiples of these are correct.) A^{2000}v⃗_{1} = 2^{2000}v⃗_{1} and A^{2000}v⃗_{2} = 2^{2000}v⃗_{2}. As any vector v⃗ is a linear combination of v⃗_{1} and v⃗_{2}, A^{2000}v⃗ = 2^{2000}v⃗. Thus

A^{2000} = 2^{2000}I = ( \begin{matrix}2^{2000} & 0 \\ 0 & 2^{2000} \end{matrix} ) .