Differential equations (function of x)
Master differential equations of the form dy/dx = f(x) for Year 12 Specialist Mathematics in Queensland (QCAA). When the right-hand side depends only on \(x\), the equation is solved by direct integration, giving a family of solution curves.
You will learn to integrate the right-hand side to find the general solution with its constant \(c\), then apply an initial condition to fix \(c\) and state the particular solution — the foundation for modelling rates of change in biology and kinematics later in Unit 4.
Theory
A differential equation of the form \(\dfrac{dy}{dx}=f(x)\) is solved by direct integration in Year 12 Specialist Mathematics (QCAA, Queensland): the right-hand side depends only on \(x\), so \(y=\displaystyle\int f(x)\,dx\). This page shows how to find the general solution (with its constant \(c\)) and the particular solution fixed by an initial condition.
A differential equation is an equation that links a function to its derivative. The simplest type met here has the form \(\dfrac{dy}{dx}=f(x)\), where the right-hand side depends on \(x\) alone. Solving it means finding every function \(y\) whose derivative equals \(f(x)\).
Because differentiating \(y\) gives \(f(x)\), reversing that step — direct integration — recovers \(y\): \(y=\displaystyle\int f(x)\,dx\). No separation of variables is needed while the right-hand side is a function of \(x\) only.
Integrating introduces an arbitrary constant, so the answer is a whole family of curves, \(y=F(x)+c\). This is the general solution: one equation standing for infinitely many parallel curves, one for each value of \(c\).
An initial condition (or boundary condition) is a known point \((x_0,y_0)\) on the required curve. Substituting it pins down a single value of \(c\), giving the particular solution — the one member of the family that passes through that point.
For \(\dfrac{dy}{dx}=f(x)\), integrate the right-hand side. The general solution carries an arbitrary constant \(c\):
Given an initial condition \(y=y_0\) when \(x=x_0\), substitute it into the general solution and solve for \(c\):
Solving \(\dfrac{dy}{dx}=f(x)\)
- Integrate the right-hand side with respect to \(x\): \(y=\displaystyle\int f(x)\,dx\). Use the standard antiderivatives (powers, \(e^{kx}\), \(\dfrac1x\), trig, inverse-trig), dividing by the coefficient of \(x\) where needed.
- Add the constant \(c\) to state the general solution \(y=F(x)+c\).
- Apply the initial condition: substitute the given point \((x_0,y_0)\) and solve the resulting equation for \(c\).
- State the particular solution by putting that value of \(c\) back into \(y=F(x)+c\); evaluate \(y\) at a required point if the question asks.
Integrate the right-hand side term by term, and add \(c\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(3x^2-4x\) |
| \(y\) | \(=\) | \(\int (3x^2-4x)\,dx\) |
| \(=\) | \(\dfrac{3x^3}{3}-\dfrac{4x^2}{2}+c\) | |
| \(=\) | \(x^3-2x^2+c\) |
The general solution is \(y=x^3-2x^2+c\).
Integrate \(e^{3x}\) (divide by the coefficient of \(x\)) for the general solution:
| \(y\) | \(=\) | \(\int e^{3x}\,dx\) |
| \(=\) | \(\dfrac{1}{3}e^{3x}+c\) |
Substitute the initial condition \((0,2)\) to find \(c\):
| \(2\) | \(=\) | \(\dfrac{1}{3}e^{0}+c\) |
| \(2\) | \(=\) | \(\dfrac{1}{3}+c\) |
| \(c\) | \(=\) | \(\dfrac{5}{3}\) |
State the particular solution:
| \(y\) | \(=\) | \(\dfrac{1}{3}e^{3x}+\dfrac{5}{3}\) |
The particular solution is \(y=\dfrac{1}{3}e^{3x}+\dfrac{5}{3}\).
Integrate: this is the \(\dfrac{1}{\sqrt{a^2-x^2}}\) form with \(a=3\):
| \(y\) | \(=\) | \(\int \dfrac{1}{\sqrt{3^2-x^2}}\,dx\) |
| \(=\) | \(\arcsin\dfrac{x}{3}+c\) |
Substitute \((0,1)\); note \(\arcsin 0=0\):
| \(1\) | \(=\) | \(\arcsin 0+c\) |
| \(1\) | \(=\) | \(0+c\) |
| \(c\) | \(=\) | \(1\) |
State the particular solution:
| \(y\) | \(=\) | \(\arcsin\dfrac{x}{3}+1\) |
The particular solution is \(y=\arcsin\dfrac{x}{3}+1\).
Integrate for the general solution:
| \(y\) | \(=\) | \(\int (4x+1)\,dx\) |
| \(=\) | \(2x^2+x+c\) |
Use \((1,4)\) to find \(c\):
| \(4\) | \(=\) | \(2(1)^2+1+c\) |
| \(4\) | \(=\) | \(3+c\) |
| \(c\) | \(=\) | \(1\) |
The particular solution is \(y=2x^2+x+1\); substitute \(x=3\):
| \(y\) | \(=\) | \(2(3)^2+3+1\) |
| \(=\) | \(18+3+1\) | |
| \(=\) | \(22\) |
When \(x=3\), \(y=22\).
Common pitfalls
Frequently asked questions
How do you solve a differential equation of the form dy/dx = f(x)?
Integrate the right-hand side with respect to \(x\): \(y=\displaystyle\int f(x)\,dx=F(x)+c\). Because the right-hand side depends only on \(x\), direct integration is all that is needed.
What is the difference between the general and particular solution?
The general solution \(y=F(x)+c\) contains the arbitrary constant and represents a whole family of curves. A particular solution uses an initial condition to fix \(c\), giving the single curve through that point.
Why do you add a constant c?
Any two antiderivatives of \(f(x)\) differ by a constant, so integrating \(\dfrac{dy}{dx}=f(x)\) determines \(y\) only up to an added constant. That is why the general solution always carries \(+c\).
What is an initial condition and how do you use it?
An initial condition is a known point \((x_0,y_0)\) on the required curve. Substitute it into the general solution \(y=F(x)+c\) and solve for \(c\); this selects the particular solution.
How do you integrate the exponential in dy/dx = e^{3x}?
Use \(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+c\). Here \(k=3\), so \(y=\dfrac{1}{3}e^{3x}+c\). Dividing by the coefficient of \(x\) is the step most often missed.
What if dy/dx depends on y instead of x?
Then direct integration does not apply. A form like \(\dfrac{dy}{dx}=g(y)\) or \(\dfrac{dy}{dx}=f(x)g(y)\) is handled by separation of variables, a separate technique later in this topic.