5.22 Differential EquationsHL ONLY

1. What Is A Differential Equation?

A differential equation is a mathematical equation that relates an unknown function $y=f(x)$ with its derivatives (such as $\dfrac{dy}{dx}$). The strict objective is to absolutely solve for the original function $y$.

In the IB HL syllabus, we systematically focus on first-order differential equations. We evaluate three primary methods:

  • Separation of Variables
  • Homogeneous Equations (using the substitution $y=vx$)
  • Integrating Factor

2. Separation of Variables

If the differential equation can be algebraically factored into entirely independent functions of $x$ and $y$, such as:

$\dfrac{dy}{dx} = f(x)g(y)$

We can safely isolate the variables on opposite sides of the equality and strictly integrate:

$\displaystyle\int \dfrac{1}{g(y)} dy = \int f(x) dx$

EXAMPLE 1

Solve the differential equation $\dfrac{dy}{dx} = 2xy$, given that $y=1$ when $x=0$.

$$ \begin{aligned} \dfrac{dy}{dx} &= 2xy \\ \dfrac{1}{y} dy &= 2x dx \\ \int \dfrac{1}{y} dy &= \int 2x dx \\ \ln|y| &= x^2 + c \end{aligned} $$ To establish the specific solution, we apply the initial condition $x=0, y=1$ $$ \ln(1) = 0^2 + c \implies c = 0 $$ Therefore, $\ln|y| = x^2$, which mathematically converts to $\mathbf{y = e^{x^2}}$.
x y 1 y = e^{x²}

3. Homogeneous Differential Equations

If the equation natively takes the form $\dfrac{dy}{dx} = f\left(\dfrac{y}{x}\right)$, we apply the substitution $y = vx$. Differentiating implicitly with respect to $x$ using the product rule seamlessly yields

$\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

This substitution reliably converts the complex homogeneous equation into a separable equation in terms of $v$ and $x$.

EXAMPLE 2

Solve the differential equation $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$.

First, we divide the numerator by the denominator to cleanly reveal the $\left(\dfrac{y}{x}\right)$ structure $$ \dfrac{dy}{dx} = \dfrac{x}{y} + \dfrac{y}{x} = \left(\dfrac{y}{x}\right)^{-1} + \left(\dfrac{y}{x}\right) $$ We systematically execute the substitution $y=vx$, thus $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ $$ \begin{aligned} v + x\dfrac{dv}{dx} &= \dfrac{1}{v} + v \\ x\dfrac{dv}{dx} &= \dfrac{1}{v} \\ v dv &= \dfrac{1}{x} dx \\ \int v dv &= \int \dfrac{1}{x} dx \\ \dfrac{v^2}{2} &= \ln|x| + c \end{aligned} $$ Replacing $v$ back with $\dfrac{y}{x}$ gives the implicit solution $$ \mathbf{\dfrac{y^2}{2x^2} = \ln|x| + c} $$

4. Intergrating Factor

For first-order linear differential equations of the strict standard form:

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

We actively calculate the integrating factor $I(x)$ defined exclusively as:

$I(x) = e^{\int P(x) dx}$

Multiplying the entire sequence by $I(x)$ compresses the left side perfectly into the exact derivative of a product: $\dfrac{d}{dx}\left[I(x)y\right]$.

EXAMPLE 3

Solve the differential equation $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$, for $x > 0$.

The mathematical equation is perfectly linear with $$ P(x) = \dfrac{2}{x} \quad \text{and} \quad Q(x) = x $$ We evaluate the integrating factor $$ I(x) = e^{\int \frac{2}{x} dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2 $$ We multiply every term in the given equation by $x^2$ $$ \begin{aligned} x^2\dfrac{dy}{dx} + 2xy &= x^3 \\ \dfrac{d}{dx}(x^2y) &= x^3 \\ x^2y &= \int x^3 dx \\ x^2y &= \dfrac{x^4}{4} + c \\ y &= \mathbf{\dfrac{x^2}{4} + \dfrac{c}{x^2}} \end{aligned} $$