CP 9.2 - General and particular solutions | Edexcel A-Level Further Mathematics

A differential equation usually describes a family of curves, while an initial condition selects a member. You will find both kinds of solution and sketch representative curves using intercepts, gradients, stationary points and limiting behaviour.

Selecting a particular solution

Two solution curves of $y'+2y=6$ start at heights $5$ and $1$ when $x=0$ . Both approach the horizontal line $y=3$ . Do they eventually meet, or can the gap shrink forever without closing? First find how the arbitrary constant identifies each curve.

You already know

For $y'+P(x)y=Q(x)$ , the integrating factor is $\mathrm{e}^{\int P\,\mathrm{d}x}$ . Integrating gives one arbitrary constant in a first-order equation. A condition such as $y=7$ when $x=0$ supplies the coordinates of one point on its solution curve.

Keep the constant while solving the differential equation, then use the condition. Substituting a point into an unfinished equation can lose the link between the entire curve and the one value you know.

General and particular solutions

A general solution describes a family of solutions using arbitrary constants. A particular solution has those constants chosen to satisfy the supplied conditions. For example, $y=3+C\mathrm{e}^{-2x}$ is a family; $y=3+4\mathrm{e}^{-2x}$ is one member. An expression still containing an undetermined $C$ is not a particular solution.

The independent variable in the condition need not be zero. Substitute its actual value into every term, including an exponential.

Worked example 1: a condition at zero

Question. Find the general solution of $y'+2y=6$ , and the particular solution for which $y=7$ when $x=0$ .

Plan. Solve using factor $\mathrm{e}^{2x}$ , then substitute the given point into the resulting family.

\begin{aligned} (\mathrm{e}^{2x}y)'&=6\mathrm{e}^{2x}&&\text{multiply by the factor}\\ \mathrm{e}^{2x}y&=3\mathrm{e}^{2x}+C&&\text{integrate}\\ y&=3+C\mathrm{e}^{-2x}&&\text{general solution}\\ 7&=3+C&&x=0,\ \mathrm{e}^0=1\\ C&=4 \end{aligned}

Why this step? The condition chooses the value of the constant; it does not replace the variable throughout the differential equation.

Answer. General: $y=3+C\mathrm{e}^{-2x}$ . Particular: $y=3+4\mathrm{e}^{-2x}$ .

Check. At $x=0$ , $y=7$ . Also $y'=-8\mathrm{e}^{-2x}$ , giving $y'+2y=6$ .

The second example changes only the horizontal coordinate of the given point from $0$ to $1$ . The same family is selected at a different location.

Worked example 2: a condition away from zero

Question. Find the general solution of $y'+2y=6$ , and the particular solution passing through $(1,7)$ .

Plan. Obtain the same general family, then retain the exponential factor when substituting $x=1$ .

\begin{aligned} (\mathrm{e}^{2x}y)'&=6\mathrm{e}^{2x}&&\text{integrating factor}\\ \mathrm{e}^{2x}y&=3\mathrm{e}^{2x}+C&&\text{integrate}\\ y&=3+C\mathrm{e}^{-2x}&&\text{general solution}\\ 7&=3+C\mathrm{e}^{-2}&&\text{substitute }(1,7)\\ C&=4\mathrm{e}^{2}&&\text{solve for the constant} \end{aligned}

Why this step? $\mathrm{e}^{-2}$ is not $1$ . Keeping it makes the curve pass through the specified point at $x=1$ .

Answer. General: $y=3+C\mathrm{e}^{-2x}$ . Particular: $y=3+4\mathrm{e}^{-2(x-1)}$ .

Check. Putting $x=1$ gives $y=7$ , and differentiating the particular solution again gives $y'+2y=6$ .

The shifted exponential in the answer makes the supplied point especially easy to check. You may instead leave the equivalent form $3+4\mathrm{e}^{2}\mathrm{e}^{-2x}$ .

Watch out

Wrong: for the point $(1,7)$ , write $7=3+C$ . That substitutes $x=0$ . Right: $7=3+C\mathrm{e}^{-2}$ . Use both coordinates of the condition.

The next condition changes the supplied height as well as its position; substitute it directly into the family.

Sketching curves around an equilibrium

For $y=3+C\mathrm{e}^{-2x}$ , the sign of $C$ tells you whether the curve approaches its limiting height from above or below.

The constant solution $y=3$ occurs when $C=0$ . It has gradient zero everywhere and satisfies the equation directly. Such a constant solution is an equilibrium solution. It is a genuine member of the family, so do not lose it by dividing by $y-3$ in another method.

For any $C$ , $y(0)=3+C$ , $y'=-2C\mathrm{e}^{-2x}$ , and $y\to3$ as $x\to\infty$ . These results are derived from the family, not supplied in the formula booklet. Use them to organise the sketch before plotting points.

Worked example 3: approaching from above

Question. Sketch the member with $C=2$ , for $x\ge0$ , showing its initial point and limiting behaviour.

Plan. Find the intercept, the sign of the derivative and the horizontal asymptote.

\begin{aligned} y&=3+2\mathrm{e}^{-2x}\\ y(0)&=5&&\text{initial height}\\ y'&=-4\mathrm{e}^{-2x}<0&&\text{strictly decreasing}\\ y-3&=2\mathrm{e}^{-2x}>0&&\text{stays above }3\\ \lim_{x\to\infty}y&=3&&\text{exponential term tends to zero} \end{aligned}

Why this step? The positive exponential term proves that the curve stays above the asymptote; drawing it crossing $y=3$ would contradict the formula.

Answer. Draw a decreasing curve from $(0,5)$ , approaching $y=3$ from above, with no stationary point.

Check. The differential equation gives the initial gradient $6-2(5)=-4$ , agreeing with the derivative.

The second example changes only the sign of $C$ . The curve now starts below the equilibrium, and the direction of motion reverses.

Worked example 4: approaching from below

Question. Sketch the member with $C=-2$ , for $x\ge0$ , showing its initial point and limiting behaviour.

Plan. Use the same three features with the negative constant.

\begin{aligned} y&=3-2\mathrm{e}^{-2x}\\ y(0)&=1&&\text{initial height}\\ y'&=4\mathrm{e}^{-2x}>0&&\text{strictly increasing}\\ y-3&=-2\mathrm{e}^{-2x}<0&&\text{stays below }3\\ \lim_{x\to\infty}y&=3&&\text{exponential term vanishes} \end{aligned}

Why this step? A negative coefficient changes the side of the equilibrium and the derivative's sign together, so the curve rises towards the same line.

Answer. Draw an increasing curve from $(0,1)$ , approaching $y=3$ from below, with no stationary point.

Check. At its initial point, the equation gives gradient $6-2(1)=4$ , as found by differentiation.

These two curves never meet at a finite $x$ . Their vertical separation is $4\mathrm{e}^{-2x}>0$ , although it tends to zero. More generally, two distinct constants give difference $(C_1-C_2)\mathrm{e}^{-2x}$ , whose sign never changes. This family can bunch together indefinitely without any two members crossing.

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CP 9.2 - General and particular solutions

Selecting a particular solution

Sketching curves around an equilibrium

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