Mathematics 9709 · Pure Mathematics 3

What a differential equation is, and how to solve one

A differential equation tells you the speed of something at every moment. $\frac{dy}{dx}$ (or $\frac{dx}{dt}$ ) is that rate.
Solving it goes backwards: from the speed to the actual journey. You do one integral on each side.
The solution is the link between the two quantities.

You can only integrate a side if it has just one variable. So first put every $y$ with $dy$ on one side and every $x$ with $dx$ on the other. Then integrate each side once. You get just one $+C$ . Find its value from the given condition before you rearrange.

The thing you are solving for is a function

A first-order differential equation has a first derivative, $\frac{dy}{dx}$ or $\frac{dx}{dt}$ , as one of its terms.
Solving a normal equation gives you a number. Solving a differential equation gives you a whole function: the link between the two variables.
$\frac{dy}{dx}=3x^2$ just asks “what has gradient $3x^2$ ?” Integrate once and you get $y=x^3+C$ .
You cannot skip that $+C$ . A whole stack of curves $y=x^3+C$ have the same gradient. So this is the general solution: the whole family at once.

General solution = the whole family (with $+C$ ). Particular solution = the one curve that a given condition picks out.

The one picture behind the whole chapter

A first-order DE does not really give you a curve. It gives you a direction at every point of the plane. $\frac{dy}{dx}=f(x,y)$ says “at the point $(x,y)$ , set off at gradient $f(x,y)$ .” Draw that gradient at many points and you get a slope field: a field of little arrows. A solution is any curve that follows the arrows all the way along. Each value of $+C$ is a curve that starts from a different point. The boundary condition tells you which one. Integrating is how you write that one curve as a formula.

Notation tip: $\dot{x}$ is short for $\frac{dx}{dt}$ . The “with respect to time” is usually left unsaid.

Separate, integrate, pin the constant

Most exam equations look like $\frac{dy}{dx}=f(x)\,g(y)$ with $x$ and $y$ mixed together.
As it stands you cannot integrate the right side over $x$ , because of the $y$ sitting in it. So split them up first.

Rearrange to $\dfrac{1}{g(y)}\,dy=f(x)\,dx$ . Put an integral sign on each side and integrate once. Write one $+C$ on the right.

Here is the rule that makes this allowed. Integrating $\frac{1}{g(y)}\frac{dy}{dx}$ over $x$ gives the same answer as integrating $\frac{1}{g(y)}$ over $y$ .
That is exactly what “ $dy$ on one side, $dx$ on the other” gives you.

Worked example

The model solve: $\dfrac{dy}{dx}=xy$ , with $y=2$ when $x=0$

This one example shows the two moves half of all paper questions need: get $\ln|y|$ from integrating $\frac1y$ , then un-log back to a clean $y=\ldots$ .

Separate. Divide by $y$ , multiply by $dx$ : $\dfrac{1}{y}\,dy=x\,dx$ .
Integrate each side once (the standard integrals from the integration chapter): $\ln|y|=\tfrac{1}{2}x^2+C$ . One constant, on the right.
Here $y>0$ , so $|y|$ is just $y$ : $\ln y=\tfrac{1}{2}x^2+C$ .
Un-log (raise $e$ to both sides): $y=e^{\frac12 x^2+C}=e^{C}\,e^{\frac12 x^2}$ . Rename the lump $e^{C}$ as a single constant $D$ .

General solution: $y=D\,e^{\frac12 x^2}$ . Now pin $D$ with $y=2$ at $x=0$ : $2=D\,e^{0}=D$ .

Answer

y=2e^{\frac12 x^2}

That is the single curve below, picked out of the whole $y=D\,e^{\frac12 x^2}$ family by the one condition.

The particular solution y = 2 e^(x²/2): one curve from the family, fixed by y = 2 at x = 0 (the gold dot).

Below is the slope field of this equation, $\frac{dy}{dx}=xy$ : each faint dash is the gradient the equation gives at that point.
Drag the gold starting point and watch the blue particular solution re-draw itself, always tangent to the dashes.
Move it onto $(0,2)$ and you get back the curve we just solved for.

x₀y₀

field dy/dx = x y·curve y = 2.00 e^((x² − 0.00)/2)

Separating and integrating just writes that one curve as a formula: the curve the field and the starting point pick out.

When the modulus drops, and renaming the constant

Integrating $\frac1y$ properly gives $\ln|y|$ .
But when the question says $y>0$ (or $V>0$ , or $0<x<300$ ), $y$ is already positive, so $|y|=y$ and the bars come off.

Keep $|y|$ unless the question tells you the sign. The moment it says $y>0$ , write $\ln y$ . And $D=e^{C}$ is just a new name for the constant after un-logging, not a second one.

Do not make up two constants either. Roll the constants from both integrals into one term on a single side.
One integral each way, one $+C$ . (Coursebook P2&3 §10.1, Key Point 10.1)

The trap that costs the most marks

Common mistake

Find the constant before you rearrange, and keep it inside the exponent. Say you rearrange to

y=\ldots

first and only then add the constant. You get

y=e^{(\ldots)}+C

, which is wrong. The constant belongs in the exponent, so the answer is

y=A\,e^{(\ldots)}

. Find

C

at the

\ln

stage, then un-log.

From here on, every later topic uses these same steps with a different thing to integrate: a trig side, an exponential side, or a fraction that first needs partial fractions.

The recipe card (textbook checklist)

Separate the variables (each side one variable).
Integrate both sides; write one arbitrary constant.
That is the general solution.
Substitute the condition to find the constant.
Check by differentiating your answer back if time allows.

(Coursebook P2&3 Ch10 checklist)

90+ aside: the extra y = 0 solution

Dividing by $g(y)$ assumes $g(y)\neq 0$ . For something like $\frac{dy}{dx}=\ldots\,y^3$ , the flat line $y=0$ is also a real solution. The separation step skips it. In 9709 papers it is almost never the marked answer, so we leave it here. (Coursebook P2&3 Worked Example 10.2)

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