Differential Equations — H2 Maths help

A differential equation relates a function to its derivative. In H2 Maths (syllabus 9758) you work exclusively with first-order separable equations — those where $\dfrac{dy}{dx}$ can be written as a product of a function of $x$ and a function of $y$ , so the variables can be separated and each side integrated independently.

What you need to know

A separable first-order ODE has the form $\dfrac{dy}{dx} = f(x)\,g(y)$ ; rewrite it as $\dfrac{1}{g(y)}\,dy = f(x)\,dx$ then integrate both sides.
The result after integration is the general solution, which contains an arbitrary constant $C$ .
A particular solution is found by substituting a given initial or boundary condition (e.g. $y = y_0$ when $x = x_0$ ) to determine $C$ .
When a quantity increases or decreases at a rate proportional to itself, the model is $\dfrac{dy}{dt} = ky$ (constant $k$ ), giving the exponential solution $y = Ae^{kt}$ .
Always include the modulus in $\ln|y|$ when integrating $\dfrac{1}{y}$ ; you may write $|y| = e^C e^{f(x)}$ and absorb $\pm e^C$ into a single constant $A$ once the context fixes the sign.
After finding a solution, verify by differentiating and checking that the original equation is satisfied.

Worked example 1 — separation of variables (general solution)

Find the general solution of $\dfrac{dy}{dx} = 3x^2 y$ .

Separate variables and integrate:

\frac{1}{y}\,dy = 3x^2\,dx

\int \frac{1}{y}\,dy = \int 3x^2\,dx \quad\Longrightarrow\quad \ln|y| = x^3 + C.

Exponentiating both sides:

|y| = e^{x^3 + C} = e^C\,e^{x^3}.

Writing $A = \pm e^C$ (an arbitrary non-zero constant):

y = A e^{x^3}.

Verification: $\dfrac{dy}{dx} = 3x^2 A e^{x^3} = 3x^2 y$ . Satisfied.

Worked example 2 — particular solution from an initial condition

Given $\dfrac{dy}{dx} = \dfrac{x}{y}$ and $y = 3$ when $x = 0$ , find the particular solution.

Separate and integrate:

y\,dy = x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + C.

Apply the initial condition $y = 3$ , $x = 0$ :

\frac{9}{2} = 0 + C \quad\Longrightarrow\quad C = \frac{9}{2}.

Particular solution:

y^2 = x^2 + 9 \quad (y > 0).

Verification: Differentiating implicitly gives $2y\,\dfrac{dy}{dx} = 2x$ , so $\dfrac{dy}{dx} = \dfrac{x}{y}$ . Satisfied.

Worked example 3 — modelling (exponential decay)

A substance decays at a rate proportional to its mass $m$ (grams) at time $t$ (minutes). Initially $m = 80$ g; after 10 minutes $m = 40$ g. Find $m$ in terms of $t$ .

The model is $\dfrac{dm}{dt} = -km$ for some $k > 0$ . Separating variables:

\frac{1}{m}\,dm = -k\,dt \quad\Longrightarrow\quad \ln m = -kt + C \quad (m > 0).

So $m = Ae^{-kt}$ . Using $m = 80$ at $t = 0$ : $A = 80$ , giving $m = 80e^{-kt}$ .

Using $m = 40$ at $t = 10$ :

40 = 80e^{-10k} \quad\Longrightarrow\quad e^{-10k} = \tfrac{1}{2} \quad\Longrightarrow\quad k = \frac{\ln 2}{10}.

m = 80\,e^{-\tfrac{t\ln 2}{10}} = 80 \cdot 2^{-t/10}.

Verification: $\dfrac{dm}{dt} = 80 \cdot \left(-\dfrac{\ln 2}{10}\right) 2^{-t/10} = -\dfrac{\ln 2}{10}\,m = -km$ . Satisfied.

Common mistakes

Forgetting $+C$ on one side only — after integrating both sides, combine into a single arbitrary constant rather than writing $C_1$ and $C_2$ separately.
Dropping the modulus in $\ln|y|$ : write $\ln|y|$ when integrating $\dfrac{1}{y}$ , then justify the sign from context before removing it.
Not verifying the solution — always differentiate your answer and substitute back to confirm it satisfies the original equation.
Applying an initial condition before integrating — find the general solution first, then substitute the condition to determine $C$ .