Difficulty: Intermediate | Prerequisites: Integration techniques, exponential and logarithmic functions
Separable ODEs are solved by moving all y-terms to one side and all x-terms to the other, then integrating both sides. Initial value problems pin down the arbitrary constant. Sequences are functions on the natural numbers, and you determine convergence by evaluating the limit as n approaches infinity, using tools like the squeeze theorem or known limit forms such as (1 + a/n)^(bn).
Separable differential equation
An ODE that can be written in the form g(y) dy = f(x) dx, so that the variables are fully separated before integrating.
In simple terms, you can shuffle all the y-stuff to one side and all the x-stuff to the other.
General solution
The family of all solutions to a differential equation, expressed with an arbitrary constant C.
Think of it as every possible curve that satisfies the equation.
Particular solution / initial value problem (IVP)
A specific solution obtained by plugging in an initial condition (e.g., y(0) = 2) to determine C.
Think of it as picking the one curve from the family that passes through a given point.
Direction field (slope field)
A grid of small line segments whose slopes equal y' at each point (x, y), giving a visual picture of how solutions behave.
In simple terms, it is a map of arrows that shows which way solutions flow.
Sequence
An ordered list of numbers a_1, a_2, a_3, ... defined by a formula a_n = f(n).
Convergent sequence
A sequence {a_n} for which lim (n -> infinity) a_n exists and is finite.
Divergent sequence
A sequence whose limit does not exist or is infinite.
Write the equation in the form dy/dx = f(x) * g(y) (or recognise it can be rearranged to that form).
Separate: (1/g(y)) dy = f(x) dx.
Integrate both sides.
Solve for y if possible; include the constant C.
Separate: (1/y) dy = (x^2 + 1) dx.
Integrate: ln|y| = x^3/3 + x + C.
Solve: y = A * e^(x^3/3 + x), where A = +/- e^C (or A = 0 for the trivial solution).
Factor: y' = y(y - 1). Separate: dy / [y(y - 1)] = dx.
Partial fractions: 1/[y(y - 1)] = 1/(y - 1) - 1/y.
Integrate: ln|y - 1| - ln|y| = x + C, so ln|(y - 1)/y| = x + C.
Solve: (y - 1)/y = A * e^x, leading to y = 1 / (1 - A * e^x).
Note the equilibrium solutions y = 0 and y = 1 (where y' = 0).
Given y = (ln(x) + K) / x, verify it satisfies x^2 y' + xy - 1 = 0.
Compute y' using the quotient rule, substitute into the equation, and confirm the left side simplifies to 0.
Solve the ODE to get the general solution (with constant C).
Plug in the initial condition to determine C.
Write the particular solution.
Separate: y^2 dy = (x - 1) dx.
Integrate: y^3/3 = x^2/2 - x + C.
Apply y(0) = 2: 8/3 = 0 - 0 + C, so C = 8/3.
Solution: y^3 = (3/2)x^2 - 3x + 8, or y = [(3/2)x^2 - 3x + 8]^(1/3).
Separate: (1/y) dy = 4/(x + 3) dx.
Integrate: ln|y| = 4 ln|x + 3| + C, so y = A(x + 3)^4.
Apply y(-2) = 1: 1 = A(1)^4, so A = 1.
Solution: y = (x + 3)^4.
Problem 15 asks you to match four ODEs to their slope fields. The strategy:
y' = x: the slope depends only on x. Every point in a vertical column has the same slope. Slopes are zero along x = 0 and grow linearly left and right.
dy/dx = 1 + x: similar to the above but shifted up by 1. Every slope in a column is the same; the zero-slope column is at x = -1.
y' = y + x: slopes depend on both x and y. Along the line y = -x (where y + x = 0), slopes are zero. The field tilts differently above and below that line.
y' - 5y = x, i.e. y' = 5y + x: the y-term dominates. Solutions blow up quickly away from equilibrium. The field shows very steep slopes when y is even slightly nonzero.
When matching, look for where the slopes are zero (the nullcline), whether the slopes depend on x only, y only, or both, and how steeply solutions diverge.
Squeeze theorem: If a_n <= b_n <= c_n and lim a_n = lim c_n = L, then lim b_n = L.
Known limit: lim (1 + a/n)^(bn) = e^(ab) as n -> infinity.
Bounded and monotone: A bounded, monotone sequence converges.
Since -1 <= sin(n) <= 1, we have -1/(50n) <= a_n <= 1/(50n).
Both bounds go to 0, so by the squeeze theorem, lim a_n = 0. Convergent.
Rewrite as [(1 + 3/(2n))^(2n)]^(5/2).
The inner expression tends to e^3 as n -> infinity.
So lim a_n = (e^3)^(5/2) = e^(15/2). Convergent.
0 <= cos^2(n) <= 1, so 0 <= a_n <= 1/2^n.
1/2^n -> 0, so by the squeeze theorem, lim a_n = 0. Convergent.
As n -> infinity, n*pi/(n + 1) -> pi.
So a_n -> cos(pi) = -1. Convergent.
Sequence: 3/16, 4/25, 5/36, 6/49, ...
Numerators: 3, 4, 5, 6, ... = n + 2.
Denominators: 16, 25, 36, 49, ... = (n + 3)^2.
General term: a_n = (n + 2) / (n + 3)^2.
Students often drop the absolute value when integrating 1/y. Remember that ln|y| is the antiderivative, and you must account for positive and negative cases when solving for y.
When separating variables, do not forget equilibrium solutions. For y' = y^2 - y, the solutions y = 0 and y = 1 satisfy y' = 0 and are missed by the separation process.
For the limit (1 + 3/(2n))^(5n), students sometimes try to apply L'Hopital's rule directly. Recognise the standard form (1 + a/n)^(bn) -> e^(ab) instead.
A constant sequence {-1, -1, -1, ...} converges (to -1). Do not confuse it with the alternating sequence {-1, 1, -1, 1, ...}, which diverges.
Direction field matching (Problem 15) is a quick-points question if you know what to look for. Check the nullcline first.
Verification problems (Problem 19) require clean differentiation and algebra. Show every substitution step.
The squeeze theorem comes up often for sequences involving sin(n), cos(n), or (-1)^n in the numerator with a growing denominator.
IVPs are almost always tested. Practise the full pipeline: separate, integrate, apply the condition, solve for y.
True or False: y' = x*y + y is separable. (True: factor as y' = y(x + 1).)
Fill in the blank: lim (1 + 5/n)^(3n) = ____. (Answer: e^15.)
True or False: The sequence a_n = (-1)^n / n converges to 0. (True: |a_n| = 1/n -> 0.)
True or False: If y' = y^2 - y, then y = 1 is an equilibrium solution. (True: y' = 1 - 1 = 0.)
Fill in the blank: On a direction field for y' = x, slopes are zero along the line x = ____. (Answer: 0.)
Q: Solve the IVP y' = (x - 1)/y^2, y(0) = 2.
A: Separate: y^2 dy = (x - 1) dx. Integrate: y^3/3 = x^2/2 - x + C. Apply y(0) = 2: C = 8/3. So y = [(3/2)x^2 - 3x + 8]^(1/3).
Q: Find the general solution of y' = (x^2 + 1)y.
A: Separate: (1/y) dy = (x^2 + 1) dx. Integrate: ln|y| = x^3/3 + x + C. So y = A * e^(x^3/3 + x).
Q: Does the sequence a_n = cos^2(n)/2^n converge? If so, find its limit.
A: Yes. 0 <= cos^2(n)/2^n <= 1/2^n, and 1/2^n -> 0. By the squeeze theorem, the limit is 0.
Q: Find the limit of the sequence (1 + 3/(2n))^(5n).
A: Rewrite as [(1 + 3/(2n))^(2n)]^(5/2). The inner part -> e^3. So the limit is e^(15/2).
Q: Show that y = (ln(x) + K)/x solves x^2 y' + xy - 1 = 0.
A: y' = (1/x * x - (ln(x) + K) * 1) / x^2 = (1 - ln(x) - K) / x^2. Then x^2 y' = 1 - ln(x) - K, and xy = ln(x) + K. Sum: 1 - ln(x) - K + ln(x) + K = 1. So x^2 y' + xy - 1 = 0.
Separable ODE, separation of variables, initial value problem, IVP, general solution, particular solution, direction field, slope field, nullcline, equilibrium solution, sequence convergence, squeeze theorem, divergent sequence, limit of a sequence, exponential limit form, Calculus II, MA-UY 1124, MA-UY 1424, NYU Courant