MATH 104, Sections 7.2 and 8.1 | Source: Rimmer lecture notes
Difficulty: Intermediate | Prerequisites: Basic integration and differentiation, u-substitution, product rule
Section 7.2 introduces the simplest class of differential equations you can actually solve by hand: separable equations. These appear constantly in modelling (population growth, radioactive decay, Newton's cooling law). Section 8.1 then opens Chapter 8's toolkit of integration techniques with integration by parts, the method you reach for when u-substitution does not work, particularly when the integrand is a product of two different types of function (a polynomial times an exponential, for example). Both topics are foundational for the rest of Calculus II.
A separable differential equation can be rewritten so that all the y terms are on one side and all the x terms are on the other, then both sides are integrated. Integration by parts reverses the product rule: you split the integrand into u and dv, then apply the formula integral of u dv = uv minus integral of v du, trading the original integral for a simpler one.
Separable differential equation
A first-order ODE dy/dx = f(x, y) where the right-hand side can be factored as g(x) times h(y). In simple terms, you can pull apart the x-stuff and the y-stuff onto opposite sides of the equation.
Initial condition
A given point (x_0, y_0) that the solution must pass through. It lets you solve for the constant of integration and get a particular solution rather than a general family.
Integration by parts (IBP)
A technique that reverses the product rule. You split the integrand into a part you differentiate (u) and a part you integrate (dv), then apply the formula: integral of u dv = uv minus integral of v du.
LIATE hierarchy
A mnemonic for choosing u in integration by parts. Functions earlier in the list make better choices for u: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Choose u from whichever type appears earliest.
Tabular integration
A shortcut for repeated integration by parts when one factor is a polynomial (which eventually differentiates to zero) and the other is easily integrated repeatedly. You build a table of successive derivatives and antiderivatives and alternate signs.
Procedure:
Start with dy/dx = f(x, y). Factor the right side as g(x) times h(y).
Rewrite as dy / h(y) = g(x) dx (separate the variables).
Integrate both sides: integral of dy / h(y) = integral of g(x) dx.
If possible, solve for y in terms of x. If an initial condition is given, use it to find the constant C.
Example 1: Exponential growth
dy/dx = ky. Separate: dy/y = k dx. Integrate: ln|y| = kx + C. Solve: y = Ae^(kx), where A = e^C.
Example 2: With an initial condition
dy/dx = y^2 x, y(1) = 6.
Separate: dy / y^2 = x dx. Integrate: -1/y = x^2/2 + C. Apply y(1) = 6: -1/6 = 1/2 + C, so C = -2/3. The particular solution is -1/y = x^2/2 minus 2/3, which gives y = 6 / (4 minus 3x^2).
Example 3: Factoring required
dy/dx = (3x^2 y^3 minus 6x^2) / y^2. Factor: dy/dx = 3x^2 (y^3 minus 2) / y^2. Separate: y^2 / (y^3 minus 2) dy = 3x^2 dx. The left side integrates with a u-substitution (u = y^3 minus 2).
Derivation from the product rule:
The product rule says d/dx [f(x) g(x)] = f'(x) g(x) + f(x) g'(x). Integrating both sides and rearranging:
\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int f'(x)\,g(x)\,dxIn shorthand notation, with u = f(x) and dv = g'(x) dx:
\int u\, dv = u\,v - \int v\, duThe big picture: You are trading one integral for another. The goal is to make the new integral (integral of v du) simpler than the original (integral of u dv).
Choosing u and dv (LIATE):
Choose u to be the function that becomes simpler when differentiated.
Make sure dv can be readily integrated.
The LIATE mnemonic ranks function types in order of preference for u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
Example: Integral of x e^(5x) dx.
Let u = x (algebraic, differentiates to 1) and dv = e^(5x) dx (exponential, integrates to e^(5x)/5). Then du = dx and v = e^(5x)/5. Applying the formula: x e^(5x)/5 minus integral of e^(5x)/5 dx = x e^(5x)/5 minus e^(5x)/25 + C.
Common wrong choice: Letting u = e^(5x) and dv = x dx gives v = x^2/2 and du = 5e^(5x) dx. The new integral, integral of (x^2/2)(5e^(5x)) dx, is harder than the original. This is how you know you picked the wrong u.
Tabular method (tic-tac-toe method):
This shortcut works when you have one of two situations:
A polynomial times an exponential or trigonometric function (the polynomial eventually differentiates to zero).
Two functions where one cycles under repeated differentiation (e.g. e^x times sin x, where you apply IBP twice and the original integral reappears).
For the first case, build a table with two columns. In the left column, write successive derivatives of the polynomial (it will reach zero). In the right column, write successive antiderivatives of the other factor. Multiply diagonally with alternating signs (+, -, +, -) and add the products.
Cycling integrals:
Sometimes applying IBP twice brings back the original integral. For example, integral of e^x sin x dx. After two rounds of IBP, you get the original integral on the right side. Collect it on the left, and solve algebraically for the integral.
Topic | Formula | When to use |
|---|---|---|
Separable DE | dy/h(y) = g(x) dx, then integrate both sides | dy/dx factors as g(x) times h(y) |
Exponential growth/decay | y = Ae^(kx) | dy/dx = ky |
Integration by parts | integral of u dv = uv minus integral of v du | Product of two different function types |
LIATE priority for u | L > I > A > T > E | Logarithmic first, Exponential last |
Tabular integration | Diagonal products with alternating signs | Polynomial times exponential or trig |
Separable differential equations model any process where the rate of change is proportional to the current quantity: radioactive decay, population growth, Newton's law of cooling, and continuously compounded interest all reduce to dy/dx = ky. Integration by parts is the workhorse behind solving differential equations that arise in electrical engineering (RC circuits), mechanical vibration analysis, and probability (computing expected values of continuous distributions).
Students sometimes try to separate an equation that is not separable. If the right side cannot be written as g(x) times h(y), the method does not apply.
When separating variables, students forget to divide by h(y) on both sides, or they drop the dx and dy, leading to nonsensical expressions.
In IBP, the most common error is choosing u and dv so that the new integral is harder. If that happens, swap your choice.
Students forget the constant of integration after integrating both sides of a separable DE. You only need one C (on either side, not both).
Separable DEs with initial conditions are a near-certain exam question. The algebra of solving for C and then solving for y often trips people up.
IBP problems typically involve ln x, x^n e^(ax), x^n sin(ax), or x^n cos(ax). Recognise these patterns on sight.
The cycling-integral technique (IBP applied twice with the original integral reappearing) is a classic exam problem. Expect integral of e^x sin x dx or integral of e^x cos x dx.
Tabular integration saves time on problems with x^2 e^x or x^3 sin x, where you would otherwise need three or four rounds of IBP.
True or False: Every first-order ODE is separable.
False. Only those where dy/dx can be factored as g(x) times h(y).
Fill in the blank: In integration by parts, you trade one integral for ______.
another (hopefully simpler) integral.
True or False: In the LIATE hierarchy, trigonometric functions should be chosen as u before algebraic functions.
False. Algebraic (A) comes before Trigonometric (T) in LIATE.
Fill in the blank: When IBP produces the original integral on the right side, you ______.
collect it on the left and solve algebraically.
Q: Solve dy/dx = ky and express y in terms of x.
A: Separate: dy/y = k dx. Integrate: ln|y| = kx + C. Exponentiate: y = Ae^(kx), where A = e^C.
Q: Evaluate the integral of x ln x dx using integration by parts.
A: Let u = ln x (L in LIATE), dv = x dx. Then du = 1/x dx, v = x^2/2. Apply the formula: (x^2/2) ln x minus integral of (x^2/2)(1/x) dx = (x^2/2) ln x minus integral of x/2 dx = (x^2/2) ln x minus x^2/4 + C.
Q: Evaluate the integral of x^2 e^(3x) dx.
A: Use tabular integration. Differentiate x^2 repeatedly: 2x, then 2, then 0. Integrate e^(3x) repeatedly: e^(3x)/3, then e^(3x)/9, then e^(3x)/27. With alternating signs: x^2 (e^(3x)/3) minus 2x (e^(3x)/9) plus 2 (e^(3x)/27) + C.
Separable DEs connect forward to the rest of differential equations in later courses (linear first-order, exact equations). Integration by parts is a prerequisite for the remaining integration techniques in Chapter 8 (trig integrals, trig substitution, partial fractions). IBP also appears in the derivation of reduction formulas, which give you general results for families of integrals like integral of sin^n x dx.
Separable differential equation, separation of variables, exponential growth, exponential decay, initial value problem, integration by parts, IBP, LIATE, tabular integration, tic-tac-toe method, cycling integral, product rule in reverse, MATH 104, Calculus II techniques, Rimmer, Penn State Math 104