First Order ODEs

What is an ODE?

An ordinary differential equation (ODE) relates an unknown function to its derivatives:

$F\left(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots\right) = 0$

Order: the highest derivative appearing.

Linear vs nonlinear: Linear means $y$ and its derivatives appear to the first power only.

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Integrable ODEs

The simplest case: $\frac{dy}{dx} = f(x)$

Solution by integration:

$y = \int f(x)\,dx + C$

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Separable ODEs

Form: $\frac{dy}{dx} = f(x)g(y)$

Method: Separate variables and integrate:

$\int \frac{dy}{g(y)} = \int f(x)\,dx$

Example: Solve $\frac{dy}{dx} = xy$.

Separate: $\frac{dy}{y} = x\,dx$

Integrate: $\ln|y| = \frac{x^2}{2} + C$

Solve: $y = Ae^{x^2/2}$ where $A = \pm e^C$.

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Linear First Order ODEs

Standard form:

$\frac{dy}{dx} + P(x)y = Q(x)$

Integrating Factor Method

Multiply by integrating factor $\mu(x) = e^{\int P(x)\,dx}$:

$\mu \frac{dy}{dx} + \mu P y = \mu Q$

The left side is $\frac{d}{dx}(\mu y)$, so:

$\frac{d}{dx}(\mu y) = \mu Q$

Integrate:

$y = \frac{1}{\mu}\int \mu Q\,dx + \frac{C}{\mu}$

Example: Solve $y' + 3y = e^{-2x}$.

$P = 3$, so $\mu = e^{3x}$.

$\frac{d}{dx}(e^{3x}y) = e^{3x}e^{-2x} = e^x$

$e^{3x}y = e^x + C$

$y = e^{-2x} + Ce^{-3x}$

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Bernoulli Equation

Form: $\frac{dy}{dx} + P(x)y = Q(x)y^n$

Method: Substitute $z = y^{1-n}$ to transform to linear.

Differentiating: $\frac{dz}{dx} = (1-n)y^{-n}\frac{dy}{dx}$

Substituting gives a linear ODE in $z$:

$\frac{dz}{dx} + (1-n)Pz = (1-n)Q$

Example: Solve $y' - xy = xy^2$.

Here $n = 2$. Let $z = y^{-1}$, so $\frac{dz}{dx} = -y^{-2}y'$.

$-y^2\frac{dz}{dx} - xy y^{-2} = x$

$-\frac{dz}{dx} - xz = x$

$\frac{dz}{dx} + xz = -x$

This is linear with $P = x$. The integrating factor $\mu = e^{x^2/2}$.

Solve: $z = -1 + Ce^{-x^2/2}$, so $y = \frac{1}{-1 + Ce^{-x^2/2}}$.

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Homogeneous Substitution

Form: $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$

Method: Let $u = \frac{y}{x}$, so $y = ux$ and $\frac{dy}{dx} = u + x\frac{du}{dx}$.

Substitute:

$u + x\frac{du}{dx} = f(u)$

$\frac{du}{dx} = \frac{f(u) - u}{x}$

This is now separable.

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Exact Equations

An equation $\frac{dy}{dx} = -\frac{P(x,y)}{Q(x,y)}$ is exact if there exists $F(x,y)$ such that:

$\frac{\partial F}{\partial x} = P \quad \text{and} \quad \frac{\partial F}{\partial y} = Q$

Test: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

Solution: The ODE becomes $dF = 0$, so $F(x,y) = C$.

Find $F$ by integrating $P$ with respect to $x$, then matching partial derivatives with $Q$.

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Existence and Uniqueness

For the initial value problem $y' = f(x,y)$, $y(x_0) = y_0$:

If $f$ and $\frac{\partial f}{\partial y}$ are continuous near $(x_0, y_0)$, then a unique solution exists locally.

Warning: Solutions may not exist for all $x$ (finite-time blow-up).

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Why These Methods Matter

• Physical systems often yield ODEs
• The classification tells you which tool to use
• Numerical methods apply when analytical solutions don’t exist
• Understanding behaviour (steady states, stability) comes from the solution structure