CP 9.5 - Complementary functions and particular integrals | Edexcel A-Level Further Mathematics

A non-zero right-hand side supplies an input to a differential equation. You will add a particular integral to the complementary function, choose trials for polynomial, exponential and trigonometric inputs, and adjust a trial when it overlaps the complementary function.

Why two pieces make the general solution

Why not solve $y''+y=1$ by integrating the right-hand side twice? That would suggest $y=x^2/2$ , but substitution gives $y''+y=1+x^2/2$ , which is not $1$ . The $y$ -term matters as much as the second derivative.

A much simpler solution is $y=1$ . Yet $y=1+\sin x$ works too, as does $y=1+7\cos x-3\sin x$ . The forcing fixes one part of the answer while allowing the equation's unforced motion to be added freely.

You already know

The homogeneous equation $y''+ay'+by=0$ is solved using its auxiliary equation. Its general solution is the complementary function, $y_{\mathrm{CF}}$ , with two arbitrary constants. Distinct, repeated and complex roots give the three forms learnt previously.

Write $L[y]=y''+ay'+by$ . Differentiation is linear, so $L[u+v]=L[u]+L[v]$ . If $L[y_{\mathrm{CF}}]=0$ and $L[y_{\mathrm{PI}}]=f(x)$ , their sum has the required right-hand side.

Particular integral

A particular integral, $y_{\mathrm{PI}}$ , is one function satisfying the non-homogeneous equation. It has no new arbitrary constants. It is different from a particular solution, which is the full solution selected by initial or boundary conditions.

The sum gives every solution, not just some solutions: subtract a chosen particular integral from any other solution and the difference satisfies the homogeneous equation.

Complementary function plus particular integral

For $y''+ay'+by=f(x)$ , with constant $a,b$ ,

y=y_{\mathrm{CF}}+y_{\mathrm{PI}}

where $y_{\mathrm{CF}}$ solves the homogeneous equation and $y_{\mathrm{PI}}$ is one solution of the full equation.

Not in the formula booklet: learn it.

There is a useful surprise here: any two solutions of the same forced equation differ by an unforced solution. The difference between two responses can be easier to understand than either response on its own.

Key idea

Find one response to the forcing, add the complete homogeneous solution, then impose conditions on the whole answer.

The two kinds of “particular” need different information.

Polynomial particular integrals

Try a general polynomial of the same degree as the forcing. Include the lower powers as well: differentiation feeds higher powers into lower ones, so the coefficients must be solved together.

For a quadratic forcing, start with $Ax^2+Bx+C$ ; for a linear forcing, start with $Ax+B$ . These trial choices are not supplied in the formula booklet. If a trial overlaps a homogeneous term, the adjustment described in the next part is needed.

Worked example 1: a linear input

Question. Find a particular integral of $y''+3y'+2y=4x+5$ .

Plan. Try $y_{\mathrm{PI}}=Ax+B$ , substitute its derivatives and compare coefficients.

\begin{aligned} y_{\mathrm{PI}}'&=A,\qquad y_{\mathrm{PI}}''=0 && \text{differentiate the trial}\\ 0+3A+2(Ax+B)&=4x+5 && \text{substitute into the equation}\\ 2A&=4,\qquad 3A+2B=5 && \text{compare powers of }x \end{aligned}

Thus $A=2$ and $B=-1/2$ .

Why include $B$ ? The constant forcing is affected by both the derivative term and the constant part of $y$ ; a trial $Ax$ alone would be too restrictive.

Answer. $y_{\mathrm{PI}}=2x-\frac12$ .

Check. $0+3(2)+2(2x-\frac12)=4x+5$ .

Worked example 2 adds only a quadratic term to the forcing. The trial therefore gains a quadratic term too.

Worked example 2: a quadratic input

Question. Find a particular integral of $y''+3y'+2y=2x^2+4x+5$ .

Plan. Try $Ax^2+Bx+C$ , then match the coefficients of $x^2$ , $x$ and the constant.

\begin{aligned} y_{\mathrm{PI}}'&=2Ax+B,\qquad y_{\mathrm{PI}}''=2A\\ 2A+3(2Ax+B)+2(Ax^2+Bx+C)&=2x^2+4x+5\\ 2A&=2 && x^2\text{ terms}\\ 6A+2B&=4 && x\text{ terms}\\ 2A+3B+2C&=5 && \text{constant terms} \end{aligned}

So $A=1$ , $B=-1$ , $C=3$ .

Why work from the highest power down? The first equation fixes $A$ , which simplifies the next equation, and then the last.

Answer. $y_{\mathrm{PI}}=x^2-x+3$ .

Check. $2+3(2x-1)+2(x^2-x+3)=2x^2+4x+5$ .

For a constant forcing, the degree-zero version of the same method uses a constant trial, provided it is not already a homogeneous solution. Do not add arbitrary integration constants to any of these particular integrals.

Watch out

Wrong: try only $Ax^2$ for a general quadratic forcing. Right: use $Ax^2+Bx+C$ ; the derivative terms generate lower powers that need balancing.

Change the leading forcing coefficient in the matched question.

Exponential forcing and overlapping trials

An exponential is preserved by differentiation, so $K\mathrm{e}^{px}$ is the natural trial for forcing $k\mathrm{e}^{px}$ . Substitution multiplies it by $p^2+ap+b$ . If that number is zero, the trial belongs to the complementary function and produces no forcing at all.

Adjusting a trial that overlaps the complementary function

Start with the usual polynomial, exponential or sine-and-cosine trial. If it contains homogeneous solution terms, multiply the complete trial by enough powers of $x$ to remove the overlap.

For forcing $k\mathrm{e}^{px}$ , use $Kx\mathrm{e}^{px}$ if $p$ is a simple auxiliary root, and $Kx^2\mathrm{e}^{px}$ if $p$ is a repeated root.

Not in the formula booklet: learn it.

The adjustment is sometimes called resonance. It is a structural issue, not a failed attempt to choose the constant: no value of $K$ can rescue a trial that is annihilated by the homogeneous operator.

Worked example 3: an exponential that does not overlap

Question. Find a particular integral of $y''-2y'+y=4\mathrm{e}^{3x}$ .

Plan. Try $K\mathrm{e}^{3x}$ ; the auxiliary root is $1$ , so this trial is independent of the complementary function.

\begin{aligned} y_{\mathrm{PI}}'&=3K\mathrm{e}^{3x},\qquad y_{\mathrm{PI}}''=9K\mathrm{e}^{3x}\\ (9-6+1)K\mathrm{e}^{3x}&=4\mathrm{e}^{3x} && \text{substitute}\\ 4K&=4 && \text{cancel the non-zero exponential} \end{aligned}

Why this trial? Differentiation changes only its coefficient, so one coefficient equation is enough.

Answer. $y_{\mathrm{PI}}=\mathrm{e}^{3x}$ .

Check. $9\mathrm{e}^{3x}-6\mathrm{e}^{3x}+\mathrm{e}^{3x}=4\mathrm{e}^{3x}$ .

Worked example 4 changes only the forcing exponent from $3$ to $1$ . It now matches the repeated auxiliary root, so two powers of $x$ are needed.

Worked example 4: a repeated-root overlap

Question. Find a particular integral of $y''-2y'+y=4\mathrm{e}^{x}$ .

Plan. Since $y_{\mathrm{CF}}=(A+Bx)\mathrm{e}^x$ , try $Kx^2\mathrm{e}^x$ , then differentiate with the product rule.

\begin{aligned} y_{\mathrm{PI}}'&=K\mathrm{e}^x(x^2+2x)\\ y_{\mathrm{PI}}''&=K\mathrm{e}^x(x^2+4x+2)\\ y_{\mathrm{PI}}''-2y_{\mathrm{PI}}'+y_{\mathrm{PI}} &=2K\mathrm{e}^x && \text{all powers of }x\text{ cancel}\\ 2K&=4 \end{aligned}

Why not just $Kx\mathrm{e}^x$ ? That is already a complementary-function term too, so it also produces zero.

Answer. $y_{\mathrm{PI}}=2x^2\mathrm{e}^x$ .

Check. Substituting $K=2$ into the displayed cancellation gives exactly $4\mathrm{e}^x$ .

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CP 9.5 - Complementary functions and particular integrals

Why two pieces make the general solution

Polynomial particular integrals

Exponential forcing and overlapping trials

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