Year 12 Maths Extension 1 (2027) Further applications of calculus

Solve dy/dx = f(x)

Q: How do you solve dy/dx = f(x)?

Integrate f of x and add a constant: y equals the integral of f of x dx plus C.

Q: How do you find the constant C?

Substitute the initial condition y at x naught equals y naught after integrating.

Q: What is the difference between general and particular solutions?

The general solution has the arbitrary C; the particular solution has C fixed by an initial condition.

Q: Which integration methods appear here?

Whatever f of x needs: standard forms, inverse-trig, logs, or substitution.

Q: Why is there a family of solutions?

Integration introduces C, so every value of C gives a valid solution curve.

20 practice questions 2 video lessons Theory + worked examples

Learn to solve first-order differential equations where the derivative depends only on x in NSW Year 12 Mathematics Extension 1. When the rate of change is a function of x alone, the solution is found by direct integration.

You will learn to integrate to recover the original function, add the constant of integration, and use an initial condition to find the particular solution — the most direct type of differential equation in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

Solving a DE of the form dy/dx = f(x) Watch
General Solution of a Differential Equation of the Form dy/dx=f(x) Watch

Theory

When $\frac{d y}{d x} = f (x)$ depends only on $x$ , integrate directly: $y = \int f (x) d x + C$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) covers the general solution and using an initial condition for the particular solution.

When $\frac{d y}{d x} = f (x)$ depends only on $x$ , the differential equation is solved by integrating directly.

This gives the general solution $y = \int f (x) d x + C$ — a family of curves. An initial condition $y (x_{0}) = y_{0}$ fixes $C$ , giving the particular solution.

Choose the integration technique to match $f (x)$ : standard forms, inverse-trig, logarithms, or substitution.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01.

The general solution: antiderivative curves differing by a vertical shift.

An initial condition selects one curve — the particular solution.

\frac{d y}{d x} = f (x) ⟹ y = \int f (x) d x + C .

dy/dx = f(x) gives y = integral f(x) dx + C

Never drop $C$ . The general solution is a family; substitute the initial condition after integrating to find $C$ , then write the particular solution.

How to solve $\frac{d y}{d x} = f (x)$

Integrate $f (x)$ , choosing the right technique for its form.
Add the constant $C$ for the general solution.
Substitute the initial condition $y (x_{0}) = y_{0}$ to solve for $C$ .
Write the particular solution.

Example 1 — General solution

Find the general solution of

\frac{d y}{d x} = 3 x^{2} + 2 x

Solution

y

=

\int (3 x^{2} + 2 x) d x = x^{3} + x^{2} + C

y = x^3 + x^2 + C

$y = x^{3} + x^{2} + C$ .

Example 2 — With a condition

Solve

\frac{d y}{d x} = 4 x - 3

given

y (1) = 2

Solution

$y = 2 x^{2} - 3 x + C$ .

2

=

2 - 3 + C \Rightarrow C = 3

y = 2x^2 - 3x + 3

$y = 2 x^{2} - 3 x + 3$ .

Example 3 — An inverse-trig form

Find the general solution of

\frac{d y}{d x} = \frac{1}{\sqrt{25 - x^{2}}}

Solution

y

=

\int \frac{d x}{\sqrt{25 - x^{2}}} = \sin^{- 1} \frac{x}{5} + C

y = arcsin(x/5) + C

$y = \sin^{- 1} \frac{x}{5} + C$ .

Example 4 — An exponential

Solve

\frac{d y}{d x} = 6 e^{3 x}

given

y (0) = 1

Solution

$y = 2 e^{3 x} + C$ .

1

=

2 + C \Rightarrow C = - 1

y = 2 e^(3x) - 1

$y = 2 e^{3 x} - 1$ .

Common pitfalls

Omitting $C$ . The general solution is a family of curves — always add the constant.

Wrong technique. Match the integral to its form: inverse-trig, logarithm, or substitution.

Applying the condition too early. Integrate first, then substitute

y (x_{0}) = y_{0}

to find

C

Dropping a coefficient. Integrating

6 e^{3 x}

gives

2 e^{3 x}

, not

6 e^{3 x}

18 e^{3 x}

Frequently asked questions

How do you solve dy/dx = f(x)?

Integrate $f (x)$ and add a constant: $y = \int f (x) d x + C$ .

How do you find the constant C?

Substitute the initial condition $y (x_{0}) = y_{0}$ after integrating.

What is the difference between general and particular solutions?

The general solution has the arbitrary $C$ ; the particular solution has $C$ fixed by an initial condition.

Which integration methods appear here?

Whatever $f (x)$ needs — standard forms, inverse-trig, logs, or substitution.

Why is there a family of solutions?

Integration introduces $C$ , so every value of $C$ gives a valid solution curve.

← Previous subtopic

Slope (direction) fields

Next subtopic →

Solve dy/dx = g(y)

More in Further applications of calculus

Solve dy/dx = f(x)

📖 Theory

How to solve dydx=f(x)

Common pitfalls

Frequently asked questions

Theory

How to solve $\frac{d y}{d x} = f (x)$