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 (NESA outcome ME1-12-05). 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, include a 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.

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Solve dy/dx = f(x)

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