PAGE 1

Linear 1st-order Eqs; Integrating Factor

\[ A(t)\frac{dy}{dt} + B(t)y = C(t) \]

\( A, B, C \) cannot contain \( y \) or \( y' \)

if \( A(t) \neq 0 \), we can put it in the standard form by dividing by \( A(t) \)

\[ \frac{dy}{dt} + p(t)y = q(t) \]

note this includes \( \frac{dy}{dt} = ay + b \)

Key Takeaway: The standard form requires a leading coefficient of 1 for the derivative \( \frac{dy}{dt} \). Identifying \( p(t) \) correctly in standard form is the crucial first step for finding the integrating factor.

Direct Calculus Method

Some can be solved by calculus (again)

for example,

\[ \underbrace{t \frac{dy}{dt} + y}_{\text{notice it's } \frac{d}{dt}(ty)} = 1 \]

product rule:\( t\frac{dy}{dt} + y \cdot 1 \)

rewrite: \[ \frac{d}{dt}(ty) = 1 \]

PAGE 2

Solving by Direct Integration

integrate with respect to \( t \)

\[ ty = \int 1 \, dt = t + c \]

Solve:

\[ y(t) = 1 + \frac{c}{t} \]

( \( t \neq 0 \) )

Motivation for Integrating Factors

in general, we may not be able to write the left side as a deriv. of a product immediately

for example,

\[ \frac{dy}{dt} + \frac{3}{t}y = \frac{1}{t} \]

look at what happens if we multiply it by \( t^3 \) (why? explain soon)

\[ \underbrace{t^3 \frac{dy}{dt} + 3t^2 y}_{\frac{d}{dt}(t^3 y)} = t^2 \]

check: \( \frac{d}{dt}(t^3 y) = t^3 \frac{dy}{dt} + 3t^2 y \)

Concept Note: Multiplying both sides by the integrating factor \( I(t) = t^3 \) transforms the sum on the left-hand side into the exact derivative of the product \( I(t)y \), allowing direct integration.

PAGE 3

Integrating Factors for Linear First-Order Equations

\[ \frac{d}{dt}(t^3 y) = t^2 \]

integrate: \( t^3 y = \frac{1}{3} t^3 + c \)

\[ y = \frac{1}{3} + \frac{c}{t^3} \quad (t \neq 0) \]

Dividing both sides by \( t^3 \) yields the explicit solution for \( y(t) \). Note the domain restriction \( t \neq 0 \) where the factor is undefined.

That \( t^3 \) we multiplied by to make the left side deriv. of a product is called an integrating factor

depends on coefficients of terms
ALWAYS can find one for any linear 1st-order eq.

Derivation for the General First-Order Linear ODE

let's derive it for

\[ y' + p(t)y = q(t) \]

goal: find \( \mu(t) \) such that when multiplied to the eq. the left side is \( \frac{d}{dt}(\mu y) \)

Multiplying the standard linear ODE by an unknown function \( \mu(t) \) gives \( \mu y' + \mu p(t) y = \mu q(t) \). By applying the product rule \( \frac{d}{dt}(\mu y) = \mu y' + \mu' y \), matching terms requires \( \mu' = \mu p(t) \).

PAGE 4

Derivation of the Integrating Factor

\( y' + py = q \)

multiply by \( \mu \):

\[ \underbrace{\mu y' + p\mu y}_{\text{want it to be } \frac{d}{dt}(\mu y) = \mu y' + \mu' y \text{ (prod. rule)}} = q\mu \]

Concept Explanation: By multiplying the standard linear equation by an unknown function \( \mu(t) \), the goal is to make the left-hand side match the result of the product rule \( \frac{d}{dt}(\mu y) = \mu y' + \mu' y \).

so,

\[ \mu y' + p\mu y = \mu y' + \mu' y \]

which means

\[ \mu' = p\mu \quad \text{or} \quad \frac{d\mu}{dt} = p(t)\mu \]

this is itself a linear eq. in \( \mu \)

this we can solve by basic calculus

\[ \frac{1}{\mu} \frac{d\mu}{dt} = p(t) \]

\( \mu \neq 0 \) otherwise \( \mu y' + p\mu y = q\mu \) is \( 0 = 0 \) (doesn't solve the eq.)

PAGE 5

Derivation of the Integrating Factor

\[ \frac{1}{\mu} \frac{d\mu}{dt} = p(t) \]

integrate with respect to \(t\)

\[ \int \frac{1}{\mu} \underbrace{\frac{d\mu}{dt} \, dt}_{d\mu} = \int p(t) \, dt \]

\[ \int \frac{1}{\mu} \, d\mu = \int p(t) \, dt \]

\[ \ln |\mu| = \int p(t) \, dt + c \]

\[ |\mu| = e^{\int p(t) \, dt + c} = \underbrace{e^c}_{\substack{\text{constant because } e \text{ and } c \\ \text{are constants}}} \cdot e^{\int p(t) \, dt} \]

\[ \mu = \underbrace{\pm e^c}_{\text{constant, call it } \text{"c"}} \cdot e^{\int p(t) \, dt} \]

\[ \mu(t) = C e^{\int p(t) \, dt} \]

ANY C would work, so for simplicity we choose \( C = 1 \)

Explanation: The integrating factor \(\mu(t)\) is used to multiply the entire linear first-order differential equation. Choosing \(C = 1\) is the simplest non-zero choice, avoiding unnecessary scalar constants across all terms.

PAGE 6

\[ \mu(t) = e^{\int p(t) \, dt} \]

Example

\[ t y' + 2y = \frac{\cos(t)}{t}, \quad y(\pi) = 0, \quad t > 0 \]

put into standard form: \( y' + p(t)y = g(t) \)

\[ y' + \underbrace{\left[ \frac{2}{t} \right]}_{p(t)} y = \underbrace{\left[ \frac{\cos(t)}{t^2} \right]}_{g(t)} \]

find integrating factor

\[ \mu(t) = e^{\int \frac{2}{t} \, dt} = e^{2\ln t} = e^{\ln t^2} = t^2 \]

Step Note: Using logarithm exponent rules, \(2 \ln(t) = \ln(t^2)\). Since exponential and logarithmic functions are inverses, \(e^{\ln(t^2)} = t^2\) for \(t > 0\).

multiply to the eq. in standard form

\[ t^2 y' + 2t y = \cos(t) \]

PAGE 7

Solving First-Order Linear ODEs: Integrating Factor Method

If \(\mu\) is correct, then left side is \(\frac{d}{dt}(\mu y)\)

& check: \[ \frac{d}{dt}(t^2 y) = t^2 y' + 2t y \quad \text{(good)} \]

Verification: Applying the product rule to \(\frac{d}{dt}(t^2 y)\) yields \(t^2 y' + 2ty\), confirming that the chosen integrating factor \(\mu(t) = t^2\) transforms the left-hand side into the exact derivative of a product.

eq. is \(\frac{d}{dt}(t^2 y) = \cos(t)\)

integrate

\[ t^2 y = \int \cos(t) \, dt = \sin(t) + C \]

\[ y(t) = \frac{\sin(t) + C}{t^2} \]

general solution (\(C\) unknown)

we were given \(y(\pi) = 0\)

\[ 0 = \frac{\sin(\pi) + C}{\pi^2} = \frac{C}{\pi^2} \implies C = 0 \]

Initial Condition: Since \(\sin(\pi) = 0\), substituting \(t = \pi\) and \(y = 0\) simplifies the numerator to \(0 + C = C\), yielding \(C = 0\).

so,

\[ y(t) = \frac{\sin(t)}{t^2} \]

particular solution

PAGE 8

Arbitrary Constant in the Integrating Factor

\[ \mu(t) = C e^{\int p(t) \, dt} \]

that \(C\) is irrelevant (\(\text{any } C \neq 0\))

why?

\[ y' + p y = q \]

multiply by \(\mu\):

\[ \cancel{C} e^{\int p(t) \, dt} y' + \cancel{C} e^{\int p(t) \, dt} p y = \cancel{C} e^{\int p(t) \, dt} q \]

Cancellation of Constant: Because the non-zero constant \(C\) multiplies every single term on both sides of the differential equation, dividing through by \(C\) eliminates it completely.

so the effect is as if we chose \(C = 1\) to begin with