I5 - Forced differential equations | AQA A-Level Further Mathematics

A nonzero right side changes what a differential equation is asking the curve to do. I5 combines the homogeneous solution with one response to polynomial, exponential or trigonometric forcing. You will choose a suitable trial, find its coefficients and recognise when overlap with the homogeneous solution requires an extra factor.

Complementary function plus particular integral

If $y''-y=2$ , can adding a constant to a homogeneous solution produce every solution of the new equation? Try $y=-2$ : its derivatives vanish and $y''-y=2$ . This gives one solution, but leaves no freedom to choose initial conditions.

Write $L[y]=y''+ay'+by$ , where $a,b$ are real constants. This is a linear operator: $L[u+v]=L[u]+L[v]$ , and constant multiples pass through it.

Complementary function and particular integral

The complementary function, $y_{\mathrm C}$ , is the general solution of the homogeneous equation $L[y]=0$ . A particular integral, $y_{\mathrm P}$ , is one function satisfying $L[y_{\mathrm P}]=f(x)$ . The general solution of $L[y]=f(x)$ is

y=y_{\mathrm C}+y_{\mathrm P}

The two arbitrary constants belong to $y_{\mathrm C}$ . This rule and the trial forms are not in the formula booklet: learn them.

Why is the sum complete? It satisfies the equation because $L[y_{\mathrm C}+y_{\mathrm P}]=0+f$ . Conversely, any solution $y$ has $L[y-y_{\mathrm P}]=f-f=0$ , so its difference from the particular integral is already included in the complementary function. Different particular integrals can therefore give the same general family.

A “particular integral” is not necessarily the final “particular solution” selected by initial conditions. Apply those conditions to the whole sum.

Worked example 1: a constant forcing term

Question. Find the general solution of $y''-y=2$ .

Plan. Solve the zero-right-side equation, then try a constant particular integral.

\begin{aligned} m^2-1&=0 && \text{homogeneous auxiliary equation}\\ y_{\mathrm C}&=Ae^x+Be^{-x}\\ y_{\mathrm P}&=C && \text{constant forcing suggests a constant trial}\\ 0-C&=2\\ C&=-2 \end{aligned}

Answer. $y=Ae^x+Be^{-x}-2$ .

Check. The two exponential terms give zero under $y''-y$ ; the constant contributes $0-(-2)=2$ . The equation holds for every $A,B$ .

Change only the forcing from the constant $2$ to the linear function $2x$ . Keep both a linear and a constant term in the trial; substitution will decide whether either coefficient vanishes.

Worked example 2: linear polynomial forcing

Question. Find the general solution of $y''-y=2x$ .

Plan. Retain the same complementary function and try $y_{\mathrm P}=Cx+D$ .

\begin{aligned} y_{\mathrm P}''-y_{\mathrm P}&=0-(Cx+D)\\ -Cx-D&=2x && \text{identity for all }x\\ C&=-2,\qquad D=0 \end{aligned}

Answer. $y=Ae^x+Be^{-x}-2x$ .

Check. Differentiating gives $y''=Ae^x+Be^{-x}$ , so subtracting $y$ gives exactly $2x$ .

For example, imposing $y(0)=0$ , $y'(0)=0$ on the second answer gives $A+B=0$ , $A-B-2=0$ . Thus $A=1,B=-1$ and $y=e^x-e^{-x}-2x$ . Omitting the derivative of the particular integral would give the wrong constants.

Watch out

Wrong: writing only the particular integral as the general solution, or inserting a third arbitrary constant into it. Add the complete two-constant complementary function; then apply conditions to the sum.

Use a changed forcing coefficient in your turn.

Polynomial, exponential and trigonometric forcing

A useful trial has a shape that differentiation preserves. Polynomials stay polynomial, an exponential retains its exponent, and sine and cosine of one frequency differentiate into one another.

For $P(m)=m^2+am+b$ , the following trials work when they do not overlap the complementary function.

Forcing	Trial particular integral	Condition or decision
Polynomial of degree $n$	Full polynomial of degree $n$	Works directly when $b\ne0$ ; include lower powers and a constant
$Ke^{kx}$ , real $k$	$Ce^{kx}$	Works directly when $P(k)\ne0$
$K\cos kx+H\sin kx$ , $k>0$	$C\cos kx+D\sin kx$	Works directly when $\mathrm{i}k$ is not an auxiliary root

These trial choices are not supplied in the formula booklet. If a listed condition fails, use the overlap adjustment in the next part. For a sum of forcing terms, find responses separately and add them, using linearity. The coefficients of a trial are fixed numbers, not new arbitrary constants.

Worked example 3: retaining the full polynomial

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

Plan. Try the full quadratic $Cx^2+Dx+E$ , including terms absent from the right side.

\begin{aligned} y_{\mathrm P}&=Cx^2+Dx+E\\ y_{\mathrm P}''+y_{\mathrm P}&=Cx^2+Dx+(E+2C)\\ C&=1,\qquad D=0,\qquad E+2C=0 \end{aligned}

Answer. $y_{\mathrm P}=x^2-2$ .

Check. Its second derivative is $2$ , so $y_{\mathrm P}''+y_{\mathrm P}=2+x^2-2=x^2$ . The constant is essential even though the forcing has no constant term.

Add only an exponential forcing term, leaving the operator and polynomial unchanged. The existing polynomial response can be kept.

Worked example 4: adding an exponential response

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

Plan. Add a trial $Fe^{2x}$ to the particular integral already found.

\begin{aligned} y_{\mathrm P}&=x^2-2+Fe^{2x}\\ y_{\mathrm P}''+y_{\mathrm P}&=x^2+(4F+F)e^{2x}\\ 5F&=1\\ F&=\frac15 \end{aligned}

Answer. $y_{\mathrm P}=x^2-2+\frac15e^{2x}$ . The full general solution would add $A\cos x+B\sin x$ .

Check. The polynomial contributes $x^2$ and the exponential contributes $(4+1)e^{2x}/5=e^{2x}$ , with no unwanted terms.

A first derivative mixes sine and cosine. Even if the forcing has just a cosine, start with both trigonometric terms.

Worked example 5: cosine forcing with a first derivative

Question. Find a particular integral of $y''+2y'+5y=5\cos x$ .

Plan. Use $y_{\mathrm P}=C\cos x+D\sin x$ and match both coefficients.

\begin{aligned} y_{\mathrm P}'&=-C\sin x+D\cos x\\ y_{\mathrm P}''&=-C\cos x-D\sin x\\ y_{\mathrm P}''+2y_{\mathrm P}'+5y_{\mathrm P} &=(4C+2D)\cos x+(4D-2C)\sin x\\ 4C+2D&=5,\qquad 4D-2C=0\\ C&=1,\qquad D=\frac12 \end{aligned}

Answer. $y_{\mathrm P}=\cos x+\frac12\sin x$ .

Check. The cosine coefficient is $4+1=5$ ; the sine coefficient is $2-2=0$ . Both match the forcing.

Change only the forcing phase, from $5\cos x$ to $5\sin x$ . The trial and coefficient equations have the same structure; the target coefficients exchange roles.

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I5 - Forced differential equations

Complementary function plus particular integral

Polynomial, exponential and trigonometric forcing

Continue this lesson.

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