Lecture 2

index

Lecture 2: First-Order ODEs

Separable ODEs

Definition (Separable ODE)

A first-order ODE $\dfrac{dy}{dx} = F(x, y)$ is separable if we can rewrite it as

F(x, y) = \frac{M(x)}{N(y)} \qquad\Longleftrightarrow\qquad \frac{dy}{dx} = \frac{M(x)}{N(y)}.

Then all $y$ -terms live on one side and all $x$ -terms on the other.

Here $F(x, y)$ depends on both variables, but factoring it as $M(x)/N(y)$ lets us integrate each side independently.

Warm-up: direct integration

For $y'(x) = 3x^2$ ,

\int dy = \int 3x^2\, dx \quad\Longrightarrow\quad y(x) = x^3 + C.

Exponential example

For $y' = y$ we separate variables:

\frac{dy}{y} = dx \quad\Longrightarrow\quad \ln|y| = x + C \quad\Longrightarrow\quad y(x) = C e^{x}.

Example with an initial value

Consider $y' = \dfrac{x}{y}$ .

y\, dy = x\, dx \quad\Rightarrow\quad \tfrac12 y^2 = \tfrac12 x^2 + C \quad\Rightarrow\quad y^2 = x^2 + C.

With $y(5) = -4$ we get $-4 = \pm\sqrt{25 + C}$ , so $C = -9$ and $y(x) = -\sqrt{x^2 - 9}$ .

Warning (Picking the right branch)

Because $y^2 = x^2 - 9$ , the solution is defined only when $|x| \ge 3$ . The initial value $x = 5$ forces us to stay on the interval $x > 3$ and to take the negative branch.

Tip (Recipe for separable ODEs)

Rewrite as $N(y)\, dy = M(x)\, dx$ , integrate both sides, and (if possible) solve for $y(x)$ . Initial conditions pick the constant and the correct sign/interval.

Which equations are separable?

$y' = xy - 2x + 4y - 8 = (x + 4)(y - 2)$ is separable: $\dfrac{dy}{y - 2} = (x + 4)\, dx$ .
$y' = y^2 + 2xy + x^2 = (x + y)^2$ is not separable directly, but with $v = x + y$ we get $v' = v^2 + 1$ , which is separable: $\arctan v = x + C$ , so $y(x) = \tan(x + C) - x$ .
$y' - 2y = x^2 e^{2x}$ looks “almost” separated, but it is not separable. It is linear, so we use integrating factors.

First-Order Linear ODEs

A first-order ODE is linear if it has the form

y' + p(x)\, y = q(x).

These need not be separable, but they all yield to the integrating factor method.

Definition (Integrating factor)

Pick $\mu(x) = e^{\int p(x)\, dx}$ . Then

\mu(x) y' + \mu(x) p(x) y = \mu(x) q(x)

collapses to $(\mu(x) y(x))' = \mu(x) q(x)$ , so

y(x) = \frac{1}{\mu(x)}\left[\int \mu(x) q(x)\, dx + C\right].

Example — $y' - 2y = x^2 e^{2x}$

$p(x) = -2$ , $q(x) = x^2 e^{2x}$ , so $\mu(x) = e^{-2x}$ .

(e^{-2x} y)' = x^2 \quad\Longrightarrow\quad e^{-2x} y = \tfrac13 x^3 + C \quad\Longrightarrow\quad y(x) = \tfrac13 x^3 + C e^{2x}.

Quick summaries

Definition (Separable ODE takeaway)

If $\dfrac{dy}{dx} = \dfrac{M(x)}{N(y)}$ , rewrite as $N(y)\, dy = M(x)\, dx$ and integrate: $\displaystyle \int N(y)\, dy = \int M(x)\, dx + C$ .

Definition (Linear ODE takeaway)

For $y' + p(x) y = q(x)$ , $\mu(x) = e^{\int p(x)\, dx}$ and

y(x) = \frac{1}{\mu(x)}\left[\int \mu(x) q(x)\, dx + C\right].

Two integrals and one constant finish the job.