MATH 304 – Ordinary Differential Equations Final Exam Solutions

I. A) By the Chain Rule, and since y = ϕ(x) solves \frac{dy}{dx} = f(y),

ψ′(x) = ϕ′(x − x_{0}) · 1 = f(ϕ(x − x_{0})) = f(ψ(x))

In addition, ψ(x_{0}) = ϕ(x − x_{0}) = ϕ(0) = α. B) The equilibria are the roots of f(x) = 0, at x = −2, 3/2 (approximately), and 4. C) The solution of the IVP with α = 1 is always decreasing since f(y) < 0 for all y with

−2 < y < 3/2. It satisfies lim_{x→∞} ϕ(x) = −2. D) The solution of the IVP with α = 2 is always increasing since f(y) > 0 for all y with

3/2 < y < 4. It satisfies lim_{x→∞} ϕ(x) = 4.

II. A) After dividing by t^{3}, the equation has the form

\frac{dz}{dt} + \frac{−4}{t}z = t^{5} sin(t)

which is first order linear. The integrating factor is e^{∫ −4/t dt} = t^{−4}, so

z = ct^{4} + t^{4} ∫ \frac{1}{t^{4}} t^{5} sin(t) dt = ct^{4} + t^{4} ∫ t sin(t) dt

Integrating by parts, we have

z = ct^{4} + t^{4}(sin(t) − t cos(t))

B) This is a homogeneous equation. Letting v = θ/s, we get \frac{dθ}{ds} = s\frac{dv}{ds} + v, so

s\frac{dv}{ds} + v = \frac{4 − 3v}{2 − v}

After some algebra:

\frac{dv}{ds} = \frac{4 − 5v + v^{2}}{2 − v} = \frac{(v − 1)(v − 4)}{2 − v}

which is separable:

∫ \frac{2 − v}{(v − 1)(v − 4)} dv = ∫ \frac{1}{s}ds

To integrate the left side, use the entries from the table of integrals in the textbook for rational functions with a factorable quadratic downstairs (b = β = 1, a = −4, α = −1):

\frac{−2}{3} ln |\frac{v − 1}{v − 4}| + \frac{1}{3}(−4 ln |v − 4| + ln |v − 1|) = ln |s| + c

This can be simplified to yield

−2 ln |v − 4| − ln |v − 1| = 3 ln |s| + c

so after exponentiating

(v − 4)^{2}(v − 1) = c/s^{3}

Substituting v = θ/s yields the final answer. C) Rearranging to

(cos(y) − e^{x})dx + (−y^{2} − x sin(y))dy = 0

the equation is seen to be exact. Solutions

x cos(y) − e^{x} − y^{3}/3 = c

III. A) The time constant is 1/k = 3/2, hence k = 2/3. Putting t = 0 at 1 am, the outside

temperature can be written as M(t) = 55 − 15 cos(πt/12). Hence by Newton’s Law of Heating and Cooling, the differential equation is:

\frac{dT}{dt} = \frac{2}{3}(55 − 15 cos(πt/12) − T)

Rearranging, we have

\frac{dT}{dt} + \frac{2}{3}T = \frac{2}{3}(55 − 15 cos(πt/12)),

which is first order linear. Using the entry from the table of integrals for ∫ e^{au} cos(nu) du, the solution is:

T = ce^{−2t/3} + e^{−2t/3} ∫ e^{2t/3} \frac{2}{3}(55 − 15 cos(πt/12)) dt

= ce^{−2t/3} + 55 − 10e^{−2t/3} ∫ e^{2t/3} cos(πt/12)) dt

= ce^{−2t/3} + 55 − \frac{10}{\frac{4}{9} + \frac{π^{2}}{144}} (\frac{2}{3} cos(πt/12) + \frac{π}{12} sin(πt/12))

≐ ce^{−2t/3} + 55 − 12.99 cos(πt/12) − 5.10 sin(πt/12)

B) After the exponential term dies off, the solution is approximately

T = 55 − 12.99 cos(πt/12) − 5.10 sin(πt/12)

Using the identity A cos(ωt) + B sin(ωt) = C cos(ωt − ϕ) where C = \sqrt{A^{2} + B^{2}} and ϕ = arctan(B/A), we see that the maximum temperature is attained when πt/12 − arctan(5.10/12.99) = π, or

t ≐ \frac{12}{π}(π + .37) ≐ 13.41

This means approximately 13.41 hours after 1am, or 2:24 pm (and every 24 hours thereafter).

IV. We have y_{2} − y_{1} = 3x and y_{1} − y_{3} = 3x^{2} − x are solutions of the homogeneous equation y′′ + py′ + q = 0. They are linearly independent because they are polynomials of two different degrees. Hence by Abel’s Identity,

Ce^{− ∫ p(x) dx} = W [3x, 3x^{2} − x] = 9x^{2}

for some constant C. Taking logarithms and differentiating, we find

p(x) = \frac{−2}{x}

V. A) y = c_{1}e^{−4x} + c_{2}xe^{−4x} B) The particular solution should have the form Ax^{2}e^{−4x} + Bx^{2} + Cx + D. Solving for

A, B, C, D yields

y_{p} = \frac{3}{2}x^{2}e^{−4x} + \frac{1}{16}x^{2} + \frac{−5}{16}x + \frac{35}{128}

C) If the 16 is replaced by 16 + ϵ, ϵ > 0, then the roots of the auxiliary equation become

r = −4 ± i\sqrt{ϵ}

The form of the homogeneous solutions changes to

y = c_{1}e^{−4x} cos(\sqrt{ϵ}x) + c_{2}e^{−4x} sin(\sqrt{ϵ}x)

and the form of the particular solution changes to Ae^{−4x} + Bx^{2} + Cx + D since e^{−4x} is no longer a solution of the homogeneous equation.

VI. A) The system has two critical points (0, 4) and (−6, −2). B) The linearization at (0, 4) has the matrix

( \begin{matrix}−1 & 1 \\ −12 & 0 \end{matrix} )

with eigenvalues λ = \frac{−1±\sqrt{−47}}{2} (complex conjugates with negative real parts). This is an inward spiral point. The linearization at (−6, −2) has the matrix

( \begin{matrix}−1 & 1 \\ −6 & 18 \end{matrix} )

with eigenvalues λ = \frac{17±\sqrt{337}}{2} (real with opposite signs). This is a saddle point. C) Solutions of linearized system at (−6, −2): Using the decimal approximations λ ≐

−.678, 17.7, we get corresponding eigenvectors:

( \begin{matrix}1 \\ .322 \end{matrix} )  ( \begin{matrix}1 \\ 18.7 \end{matrix} )

so the solutions are

c_{1}e^{−.678t} ( \begin{matrix}1 \\ .322 \end{matrix} ) + c_{2}e^{17.7t} ( \begin{matrix}1 \\ 18.7 \end{matrix} )