MATH 215/255 PRACTICE MIDTERM I

Problem I. (Short answer questions.) (10 points)

(a) (2 points) Find an integrating factor for the equation:

(t2 + 1)y′ − 2ty = t2 + 1.

The integrating factor is:

µ(t) = e∫− 2t

t2+1dt

= e− ln (t2+1) = (t2 + 1)−1.

(b) (2 points) Find the general solution of the equation:

y′′ + 4y′ + 4y = 0

The characteristic equation is: r2 + 4r + 4 = (r + 2)2, so there is a double root: r = −2. Thegeneral solution is: y = (C1 + C2t) e

−2t.

(c) (2 points) Consider the differential equation: y′ = (y− 1)(ey − 1). Sketch the direction field. If0 < y(0) < 1, then find the limit of the solution y(t) as t → ∞, without solving the equation.

Since (y − 1)(ey − 1) < 0 for 0 < y < 1 the solution y = y(t) is decreasing and approaching thestable fixed point: y∗ = 0, so y(t) → 0 if 0 < y(0) < 1.

(d) (2 points) Solve the initial value problem:

y′ = y2, y(0) = 1/2

and find the largest interval on which the solution exists.

The solution of this separable equation is: y = − 1t−2 , thus it is not defined at t = 2. The largest

”interval” is the half-line: t < 2.

(e) (2 points) Find all values of α such that all solutions of the equation tend to 0 as t → ∞.

u′′ + 2(α− 1)u′ + u = 0

The characteristic equation is: r2+2(α−1)r+1 = 0. The roots are: r1,2 = −(α−1)±√

(α− 1)2 − 1.1

2 MATH 215/255 PRACTICE MIDTERM I

If α > 1 then we have two negative real roots, or two complex roots with negative real part, so inboth cases all solutions tend to 0 as t → ∞. The answer is α > 1. (This will be more obvious afterthe spring-mass application as the positive coefficient of u′ corresponds to ”damping”.)

Problem II. (6 points) Consider the autonomous equation depending on the parameter k.

y′ = 2y − y2 − k

(i) (3 points) Find the equilibrium points for k = 0 classify them as stable or unstable and sketcha few typical solution curves.

For k = 0, we have: y′ = 2y − y2 = y(2 − y). The equilibrium points are: y1 = 0, y2 = 2. Herey1 = 0 is unstable, while y2 = 2 is stable (this is the logistic equation).

(ii) (3 points) Find the equilibrium point(s) for k = 1 and sketch a few typical solution curves.

The equation is: y′ = 2y − y2 − 1 = −(y − 1)2. the equilibrium point is: y1 = 1. Note that y′ < 0both if y > 1 or y < 1. So y(t) → 1 if y(0) > 1 and y(t) → −∞ if y(0) < 1. The equilibrium pointis semi-stable.

Problem III. (8 points) Solve the following initial value problems. Indicate the largest intervalon which the solution exists.

(i) (4 points)

(ln(xy) + x2 + y2 + (x

y+ 2yx)y′ = 0, y(1) = 1

This is an exact equation with: M(x, y) = ln(xy) + x2 + y2 and N(x, y) = xy + 2yx. Indeed:

M ′y = 1/y + 2y = N ′

x.

The solutions are of the form: H(x, y) = C, where H ′x = M and H ′

y = N . Thus

H(x, y) =

∫(x

y+ 2yx) dy = x ln y + y2x+ h(x).

From the equation H ′x = M , we get:

ln y + y2 + h′(x) = ln(xy) + x2 + y2 = ln x+ ln y + y2 + x2, so

h′(x) = ln x+ x2, ⇒ h(x) =

∫(ln x+ x2) dx = x ln x− x+

1

3x3.

MATH 215/255 PRACTICE MIDTERM I 3

The general solution (in implicit form) is:

x ln y + y2x+ x ln x− x+1

3x3 = C.

(ii) (4 points)

x2 y′ = y2 + yx+ x2, y(1) = 0

This is a homogeneous equation: y′ = (y/x)2 + (y/x) + 1. Let u = y/x, so y = ux and theny′ = u′x+ u. Substituting this into the equation gives:

u′x+ u = u2 + u+ 1 ⇒ u′x = u2 + 1.

By separating the variables, we have:∫du

u2 + 1du =

∫dx

x,

tan−1(u) = ln x+ C ⇒ u = tan (ln x+ C) ⇒ y = x tan (ln x+ C).

Problem IV. (8 points) A hot coffee of initial temperature of 80 (F) is placed on a table in roomof temperature 60 (F). Suppose that the coffee starts cooling down at a rate 2 (F/min) whiletemperature of the room is increasing at a constant rate 1 (F/min).

(b) (3 points) Write down a differential equation for the temperature of the coffee.

Let T denote the temperature of the coffee, and Tm = 60 + t the temperature of the room. ByNewton’s law of cooling:

T ′ = −k(T − 60− t).

The initial data gives: T (0) = 80 and T ′(0) = −2 (the rate of initial cooling). Thus: −2 =−k(80− 60), so k = 1/10. Then we have the initial value problem:

T ′ +1

10T = 6 +

t

10, T (0) = 80.

which is an easy linear DE to solve.

(b) (5 points) Solve the equation and calculate the the temperature of the coffee after t minutes.

Problem V. (8 points) Consider the differential equation

t2y′′ + ty′ − y = 0.

(i) (6 points) Find two solutions of the form: y = tr and verify that they form a fundamental setof solutions by calculating their Wronskian. Write down the general solution.

4 MATH 215/255 PRACTICE MIDTERM I

If y = tr, then y′ = rtr−1 and y′′ = r(r − 1)tr−2. Plugging this into the equation gives:

r(r − 1)tr + rtr − tr = 0.

Dividing both sides by tr:

r(r − 1) + r − 1 = r2 − 1 = 0

, so r = 1 or r = −1. This gives the solutions: y1 = t, y2 = t−1.

The Wronskian is:

W (y1, y2) = t−1 + t t−2 = 2t−1 ̸= 0.

The general solution is: y = C1t+ C2 t−1.

(i3) (2 points) Find the solution satisfying the initial conditions: y(1) = 1, y′(1) = 0

y(1) = C1 + C2 = 1y”(1) = C1 − C2 = 0Hence C1 = C2 = 1/2.