Differential equations dy/dx = f(x)
Master differential equations of the form \(\dfrac{dy}{dx}=f(x)\) in Year 12 VCE Specialist Mathematics. When the right-hand side depends only on \(x\), the equation is solved by direct integration, giving a family of solution curves. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Differential equations topic of Unit 4.
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: 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.