2.6 - Finding particular integrals | Cambridge A-Level Further Mathematics

The right-hand side of a linear differential equation suggests the shape of a particular integral. You will determine coefficients for polynomial, exponential and trigonometric trials, then use supplied forms when an ordinary trial overlaps the complementary function or needs extra terms.

Polynomial forcing

Can the shape of the right-hand side save us from integrating a differential equation? For $y''+3y'+2y=4x+7$ , try a straight line: differentiation cannot produce any higher powers, and the $2y$ term can reproduce the linear term.

The right-hand side is sometimes called the forcing term. For a polynomial forcing term, try a polynomial of the same degree when that is a suitable form. Include all lower powers: their coefficients may be needed to cancel terms produced by differentiation.

You already know

The general solution is the complementary function plus one particular integral. For $y''+3y'+2y=f(x)$ , the complementary function is $C_1e^{-x}+C_2e^{-2x}$ .

Use new lower-case coefficients in the trial so that they do not become confused with the arbitrary constants in the complementary function.

Worked example 1: a linear polynomial

Question. Find the general solution of $y''+3y'+2y=4x+7$ .

Plan. Try $y_p=ax+b$ , substitute and compare coefficients of $x$ and the constant.

Here $y_p'=a$ and $y_p''=0$ , so

2ax+(3a+2b)=4x+7.

Thus $2a=4$ , giving $a=2$ , and $6+2b=7$ , giving $b=1/2$ .

Answer. $y=C_1e^{-x}+C_2e^{-2x}+2x+1/2$ .

Check. For $y_p=2x+1/2$ , the left-hand side is $0+6+4x+1=4x+7$ . Each exponential in the complementary function contributes zero separately.

Change only the degree of the forcing polynomial. The trial now needs a quadratic term and both lower powers.

Worked example 2: a quadratic polynomial

Question. Find the general solution of $y''+3y'+2y=4x^2+7$ .

Plan. Use $y_p=ax^2+bx+c$ , then work down from the highest power.

Since $y_p'=2ax+b$ and $y_p''=2a$ , comparison gives

\begin{aligned} 2a&=4 \\ 6a+2b&=0 \\ 2a+3b+2c&=7. \end{aligned}

Therefore $a=2$ , $b=-6$ and $c=21/2$ . The linear coefficient on the right is zero, but a linear term in the trial is still necessary.

Answer. $y=C_1e^{-x}+C_2e^{-2x}+2x^2-6x+21/2$ .

Check. The trial gives $y_p'=4x-6$ , $y_p''=4$ , and hence

4+3(4x-6)+2(2x^2-6x+21/2)=4x^2+7.

If a proposed trial belongs to the complementary function, substitution gives zero instead of the required forcing. The form then needs adjustment; a later part uses explicitly supplied adjusted trials.

Watch out

For quadratic forcing, trying only $ax^2$ usually fails. Include $bx+c$ before comparing coefficients, even if the forcing has no linear or constant term.

Find just the particular integral in the next problem.

Exponential forcing

An exponential forcing term suggests an exponential trial because all its derivatives retain the same exponential factor.

For $ay''+by'+cy=Ke^{rx}$ , trying $y_p=ke^{rx}$ gives $k(ar^2+br+c)e^{rx}$ . If the bracket is nonzero, comparing coefficients determines $k$ . The same reasoning works for a first-order equation: in $y'+2y=8e^{2x}$ , the trial $ke^{2x}$ gives $4k=8$ .

Exponential trial

For a constant-coefficient linear equation, try $y_p=ke^{rx}$ for forcing $Ke^{rx}$ , provided that exponential is not a homogeneous solution. Substitute its derivatives to determine $k$ .

If it overlaps the complementary function, the trial must change. Check a supplied adjusted form by differentiation. This method is not in the formula booklet: learn it.

The overlap test is a practical check, not an instruction to divide by zero and continue.

Worked example 3: no overlap

Question. Find the general solution of $y''-y'-2y=9e^{3x}$ .

Plan. Find the complementary function, then try $ke^{3x}$ .

The auxiliary equation is $(m-2)(m+1)=0$ , giving $y_c=C_1e^{2x}+C_2e^{-x}$ . Substitution of the trial gives

(9k-3k-2k)e^{3x}=9e^{3x},

so $4k=9$ .

Answer. $y=C_1e^{2x}+C_2e^{-x}+\tfrac94e^{3x}$ .

Check. The particular term has first and second derivatives $\tfrac{27}4e^{3x}$ and $\tfrac{81}4e^{3x}$ . Their combination is $(81-27-18)e^{3x}/4=9e^{3x}$ .

Change only the forcing exponent from $3$ to $2$ . The ordinary exponential trial now lies inside the complementary function, so an adjusted form is supplied.

Worked example 4: a supplied trial avoids overlap

Question. Find the general solution of $y''-y'-2y=9e^{2x}$ , using the particular-integral form $y_p=kxe^{2x}$ .

Plan. Retain the preceding complementary function and differentiate the supplied product carefully.

\begin{aligned} y_p'&=k(1+2x)e^{2x} \\ y_p''&=k(4+4x)e^{2x} \\ y_p''-y_p'-2y_p&=3ke^{2x}. \end{aligned}

Thus $3k=9$ , giving $k=3$ .

Answer. $y=C_1e^{2x}+C_2e^{-x}+3xe^{2x}$ .

Check. Put $y=e^{2x}v$ directly in the original equation. The left-hand side becomes $e^{2x}(v''+3v')$ . For the particular choice $v=3x$ , this is $9e^{2x}$ , confirming the adjusted trial independently.

Trying $ke^{2x}$ instead would give zero for every $k$ ; no value of its coefficient could make it a particular integral here.

Watch out

Before cancelling an exponential and solving for its coefficient, check whether the coefficient multiplier is zero. An overlap requires a new trial, not an infinite coefficient.

Use the provided adjusted form with a new forcing size.

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2.6 - Finding particular integrals

Polynomial forcing

Exponential forcing

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