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2.1 Linear Diff. Eqs.; Method of Integrating Factors

1st-order linear eqs:

\[ P(t) \frac{dy}{dt} + Q(t)y = G(t) \]
Note: \( P(t) \), \( Q(t) \), and \( G(t) \) cannot contain \( y \).

if \( P(t) \neq 0 \), we can make the coefficient 1

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

"Standard form"

Note: \( p(t) \) and \( g(t) \) have no \( y \).

Some can be solved with calculus

for example,

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

notice this is

\[ \frac{d}{dt}(ty) = t \frac{dy}{dt} + y \cdot 1 \]

product rule

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so, eq. becomes

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

integrate with respect to \( t \)

\[ ty = t + c \]

solve for \( y \)

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

unfortunately, the left side doesn't usually directly turn into the deriv. of a product

for example,

\[ t \frac{dy}{dt} + 3y = 1 \]
left side is NOT deriv. of a product (try it!)
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rewrite:

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

multiply both sides by \( t^3 \) (why? explained soon)

\[ t^3 \frac{dy}{dt} + 3t^2 y = t^2 \]

The left side is the derivative of a product:

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

integrate

\[ t^3 y = \frac{1}{3} t^3 + C \]\[ y = \frac{1}{3} + \frac{C}{t^3} \]

that \( t^3 \) we multiplied by is called the integrating factor

(different for every equation)

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How to find the integrating factor?

First, make sure leading coefficient is 1.

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

multiply by the integrating factor \( \mu(t) \) such that the left side is the derivative of \( \mu(t) \) times \( y \)

\[ \mu \frac{dy}{dt} + p \mu y = \mu g \]

turn into

\[ \frac{d}{dt} (\mu y) = \mu \frac{dy}{dt} + \frac{d\mu}{dt} y \]

so,

\[ \mu \frac{dy}{dt} + p \mu y = \mu \frac{dy}{dt} + \frac{d\mu}{dt} y \]

equal

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So, \[ \frac{d\mu}{dt} = p\mu \] this is a diff. eq. solve for \( \mu(t) \)

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

from calculus, left side is \( \frac{d}{dt} \ln|\mu| \)

\[ \frac{d}{dt} \ln|\mu| = p \]

integrate

\[ \ln|\mu| = \int p \, dt + C \]
\[ |\mu| = e^{\int p(t) \, dt + C} = e^{\int p(t) \, dt} \cdot e^C \]
\[ \mu = \underbrace{\pm e^C}_{= C} \cdot e^{\int p \, dt} \]
\[ = C e^{\int p \, dt} \]

choose ANY \( C \neq 0 \), so choose \( C = 1 \)

\[ \mu(t) = e^{\int p(t) \, dt} \]
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How to solve \( y' + p(t)y = g(t) \)

  1. find integrating factor \( \mu(t) = e^{\int p(t) \, dt} \) works only if leading coeff = 1
  2. multiply both sides of \( y' + p(t)y = g(t) \) by \( \mu(t) \)
  3. turn left side into \( \frac{d}{dt} (\mu y) \)
  4. integrate and solve

example :

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

make leading coeff. 1

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

Find Integrating Factor

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

Simplify \( \mu \) ALL THE WAY

Multiply both sides of \( y' + \frac{2}{t} y = \frac{\cos(t)}{t^2} \) by \( \mu \)

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

If \( \mu \) is correct, the left side should turn into:

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

Check: \( t^2 y' + 2t y \)

Now diff. eq. becomes

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

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

General Solution

(infinitely-many solutions because of unknown \( C \))

We were given \( y(\pi) = 0 \). Use it to find \( C \).

Plug in \( t = \pi \), \( y = 0 \), solve for \( C \):

\[ 0 = \frac{\sin(\pi)}{\pi^2} + \frac{C}{\pi^2} \]\[ = 0 + \frac{C}{\pi^2} \quad \text{so, } C = 0 \]

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

Particular Solution

(\( C \) is known)