Nonhomogeneous Linear Equations, multivariable and vector calculus lesson

Push a swing at regular intervals and its motion is no longer set only by the swing itself. The equation gains a right side, a prescribed input that you choose, while the motion it produces is the unknown. All the work on homogeneous equations still counts. You only need one solution that answers the input, and everything else comes from the homogeneous equation you already know how to solve.

Key terms

Forcing and the linear operator

Definition A second-order linear equation in standard form is

y'' + p(t)\,y' + q(t)\,y = g(t),

where $p$ , $q$ and $g$ are given functions on an interval $I$ . Write $L[y] = y'' + p\,y' + q\,y$ for every twice differentiable $y$ on $I$ . The operator $L$ is linear, so $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]$ for all constants $c_1, c_2$ . The given function $g$ is the forcing term. The equation is homogeneous when $g = 0$ on $I$ and nonhomogeneous otherwise.

In plain words The operator $L$ is a machine that takes a candidate function and returns the forcing it would need. Solving $L[y] = g$ means finding the functions that the machine sends to the prescribed input $g$ .

Example For $L[y] = y'' - y$ and the candidate $y = t^2$ the machine returns

L[t^2] = 2 - t^2.

So $y = t^2$ solves $y'' - y = 2 - t^2$ , but it does not solve $y'' - y = 2$ .

Coefficient continuity assumptions

Definition Existence and uniqueness theorem. Suppose $p$ , $q$ and $g$ are continuous on an open interval $I$ and $t_0$ lies in $I$ . Then for any real numbers $y_0$ and $y_1$ the problem

\begin{gathered} y'' + p\,y' + q\,y = g, \\[0.4em] y(t_0) = y_0, \quad y'(t_0) = y_1 \end{gathered}

has exactly one solution, and that solution exists on all of $I$ . An equation written as $P(t)y'' + Q(t)y' + R(t)y = G(t)$ must first be divided by $P$ , which is allowed only on an interval where $P(t) \ne 0$ . Then $p = Q/P$ , $q = R/P$ and $g = G/P$ .

In plain words The theorem works on intervals where the coefficients and the forcing are continuous, and only after the equation is in standard form. Points where the leading coefficient vanishes split the line into separate intervals, and a solution formula is claimed only on one of them.

Example The equation $t^2 y'' - 2t\,y' + 2y = t^3$ has leading coefficient $t^2$ , which is zero at $t = 0$ . In standard form it reads

y'' - \frac{2}{t}\,y' + \frac{2}{t^2}\,y = t,

with coefficients continuous on $(0, \infty)$ and on $(-\infty, 0)$ . An initial condition at $t_0 = 1$ gives a unique solution on $(0, \infty)$ , and the theorem says nothing past $t = 0$ .

Complementary and particular solutions

Definition Let $p$ , $q$ and $g$ be continuous on an open interval $I$ , and let $y_1$ and $y_2$ be solutions of $L[y] = 0$ on $I$ with Wronskian $W(y_1, y_2)(t) \ne 0$ on $I$ . Any single solution $y_p$ of $L[y] = g$ on $I$ is called a particular solution. The general solution of the homogeneous equation,

y_c = c_1 y_1 + c_2 y_2,

is called the complementary solution. Theorem. Every solution of $L[y] = g$ on $I$ has the form

y = c_1 y_1 + c_2 y_2 + y_p

for exactly one pair of constants $c_1, c_2$ . The proof is one line. If $y$ solves $L[y] = g$ , then $L[y - y_p] = g - g = 0$ , so $y - y_p$ is a homogeneous solution.

In plain words The particular solution answers the forcing. The complementary part adds nothing to the forcing, because $L$ sends it to zero, so it is free to adjust the solution to the initial conditions.

Example For $y'' - y = 2$ the constant $y_p = -2$ works, since $0 - (-2) = 2$ . The homogeneous equation has $y_1 = e^t$ and $y_2 = e^{-t}$ , so every solution on $(-\infty, \infty)$ is

y = c_1 e^t + c_2 e^{-t} - 2.

Check the roles with the operator. Applying $L$ to $c_1 e^t + c_2 e^{-t}$ gives $0$ and applying it to $-2$ gives $2$ , so their sum is sent to exactly $2$ , whatever $c_1$ and $c_2$ are.

Undetermined coefficients

Definition The method applies to $a y'' + b y' + c y = g(t)$ with real constants $a \ne 0$ , $b$ , $c$ , when $g$ is a sum of terms of the form

\begin{gathered} P_m(t)\,e^{\alpha t}\cos \beta t \\[0.4em] \text{or} \quad P_m(t)\,e^{\alpha t}\sin \beta t, \end{gathered}

where $P_m$ is a polynomial of degree $m$ and $\alpha$ , $\beta$ are real constants. For one such term, try

\begin{aligned} y_p = t^s e^{\alpha t}\big(&A(t)\cos \beta t \\[0.4em] &+ B(t)\sin \beta t\big), \end{aligned}

where $A$ and $B$ are polynomials of degree $m$ with unknown coefficients, and $s$ is the multiplicity of $\alpha + i\beta$ as a root of the characteristic equation $a r^2 + b r + c = 0$ , with $s = 0$ when it is not a root. So $s$ is $0$ , $1$ or $2$ , and $s = 2$ is possible only when $\beta = 0$ . Both $A$ and $B$ are needed even when $g$ contains only a cosine or only a sine. Substitute and match coefficients. For a sum of such terms, find a particular solution for each term and add them, which is allowed because $L$ is linear.

In plain words Derivatives of polynomials, exponentials, sines and cosines stay in the same family, so a trial of the same shape as the forcing can be tuned to fit it exactly. When $s = 0$ the trial simply copies the shape of $g$ .

Example Take $y'' + 3y' + 2y = 4t^2$ . The characteristic roots are $-1$ and $-2$ , and $0$ is not one of them, so $s = 0$ and the trial is $y_p = At^2 + Bt + C$ . Substituting gives

\begin{aligned} &2A t^2 + (6A + 2B)t \\[0.4em] &\quad + (2A + 3B + 2C) = 4t^2. \end{aligned}

Matching coefficients gives $A = 2$ , then $B = -6$ , then $C = 7$ , so $y_p = 2t^2 - 6t + 7$ . With exponential forcing the same idea is even quicker. For $y'' + y = 5e^{2t}$ the trial $Ae^{2t}$ gives $4A + A = 5$ , so $y_p = e^{2t}$ .

Resonance adjustments

Definition When the forcing contains $e^{\alpha t}\cos\beta t$ or $e^{\alpha t}\sin\beta t$ and $\alpha + i\beta$ is a root of the characteristic equation, the unadjusted trial contains homogeneous solutions, which $L$ sends to $0$ , so no choice of coefficients can match $g$ . Multiplying the trial by $t^s$ , with $s$ the multiplicity of the root as above, removes every homogeneous solution from the trial and the method then succeeds. Always compare the trial with $y_c$ before substituting.

In plain words If the input already has the shape of a natural response, the plain guess is invisible to the equation. Multiply by $t$ , or by $t^2$ for a double root, to get a guess the equation can see.

Example For $y'' - y' - 2y = 3e^{2t}$ the characteristic equation $r^2 - r - 2 = (r - 2)(r + 1)$ has the simple root $r = 2$ . The trial $Ae^{2t}$ gives $0$ , so use $y_p = At\,e^{2t}$ . Its derivatives are

\begin{aligned} y_p' &= A e^{2t}(1 + 2t), \\[0.4em] y_p'' &= A e^{2t}(4 + 4t), \end{aligned}

and substituting gives $A e^{2t}(4 + 4t - 1 - 2t - 2t) = 3Ae^{2t}$ . So $A = 1$ and $y_p = t\,e^{2t}$ . The same rule handles a double root and a sinusoid.

Equation	Root to check	Particular solution
$y'' - 2y' + y = e^t$	$r = 1$ is double	$\tfrac{1}{2}t^2 e^t$
$y'' + 4y = \cos 2t$	$r = 2i$ is simple	$\tfrac{1}{4}t \sin 2t$

In the second row the oscillation has amplitude $t/4$ , which grows without bound. This is resonance.

Variation of parameters

Definition Let $p$ , $q$ and $g$ be continuous on an open interval $I$ , and let $y_1$ , $y_2$ be solutions of $y'' + p\,y' + q\,y = 0$ on $I$ with Wronskian

W = y_1 y_2' - y_1' y_2 \ne 0 \quad \text{on } I.

Look for $y_p = u_1 y_1 + u_2 y_2$ and impose $u_1' y_1 + u_2' y_2 = 0$ . Substituting into the equation then leaves $u_1' y_1' + u_2' y_2' = g$ . Solving this linear system with Cramer’s rule gives

u_1' = -\frac{y_2\,g}{W}, \qquad u_2' = \frac{y_1\,g}{W},

and any antiderivatives $u_1$ , $u_2$ on $I$ give a particular solution

y_p = -y_1 \int \frac{y_2\,g}{W}\,dt + y_2 \int \frac{y_1\,g}{W}\,dt.

The formula needs the equation in standard form with leading coefficient $1$ , and it works for any continuous $g$ , also with variable coefficients.

In plain words Take the homogeneous solution and let its constants vary with $t$ . The extra condition keeps the second derivative simple, and the Wronskian being nonzero guarantees you can always solve for $u_1'$ and $u_2'$ .

Example Solve $y'' + y = \sec t$ on $I = (-\pi/2, \pi/2)$ , where $\sec t$ is continuous. Use $y_1 = \cos t$ and $y_2 = \sin t$ with $W = 1$ . Then $u_1' = -\sin t \sec t = -\tan t$ and $u_2' = \cos t \sec t = 1$ . Since $\cos t > 0$ on $I$ , take $u_1 = \ln(\cos t)$ and $u_2 = t$ . This gives

y_p = \cos t \,\ln(\cos t) + t \sin t.

Undetermined coefficients cannot reach this answer, because $\sec t$ is not of the allowed form.

Variable coefficient forcing

Definition For $P(t)y'' + Q(t)y' + R(t)y = G(t)$ on an open interval where $P(t) \ne 0$ and all four functions are continuous, the forcing that enters variation of parameters is $g = G/P$ , not $G$ . The homogeneous solutions $y_1$ , $y_2$ must be known separately, and the result is valid only on that interval.

In plain words Divide by the leading coefficient before you use the formula. Forgetting this is the most common error with variable coefficients.

Example Take $t^2 y'' - 2t\,y' + 2y = t^3$ on $(0, \infty)$ . The functions $y_1 = t$ and $y_2 = t^2$ solve the homogeneous equation, as direct substitution shows, and $W = t \cdot 2t - 1 \cdot t^2 = t^2 \ne 0$ there. The standard form has $g = t$ . Then

\begin{aligned} u_1' &= -\frac{t^2 \cdot t}{t^2} = -t, \\[0.4em] u_2' &= \frac{t \cdot t}{t^2} = 1, \end{aligned}

so $u_1 = -\tfrac{1}{2}t^2$ , $u_2 = t$ and

y_p = -\tfrac{1}{2}t^3 + t^3 = \tfrac{1}{2}t^3.

Check it in the original equation. The left side becomes $t^2 \cdot 3t - 2t \cdot \tfrac{3}{2}t^2 + t^3 = t^3$ , as required. Using $G = t^3$ instead of $g = t$ would have produced a wrong answer.

Check yourself

Find a particular solution of $y'' - 4y = 8e^{2t}$ .

Show the answer

The root $r = 2$ of $r^2 - 4 = 0$ is simple, so try $y_p = At\,e^{2t}$ . Substituting gives $A e^{2t}(4 + 4t) - 4At\,e^{2t} = 4Ae^{2t}$ , so $A = 2$ and $y_p = 2t\,e^{2t}$ .

Solve $y'' + y = 2t$ with $y(0) = 0$ and $y'(0) = 0$ .

Show the answer

A particular solution is $y_p = 2t$ , so $y = c_1 \cos t + c_2 \sin t + 2t$ . The condition $y(0) = 0$ gives $c_1 = 0$ , and $y'(0) = c_2 + 2 = 0$ gives $c_2 = -2$ . The solution is $y = 2t - 2\sin t$ for all $t$ .

Use variation of parameters to find a particular solution of $y'' - 2y' + y = e^t/t$ on $(0, \infty)$ .

Show the answer

Here $y_1 = e^t$ , $y_2 = t e^t$ and $W = e^{2t}$ . Then $u_1' = -1$ and $u_2' = 1/t$ , so $u_1 = -t$ and $u_2 = \ln t$ , which needs $t > 0$ . This gives $y_p = -t e^t + t e^t \ln t$ . The term $-te^t$ is a homogeneous solution and may be dropped, so $y_p = t e^t \ln t$ also works.

Nonhomogeneous Linear Equations

Key terms

Forcing and the linear operator

Coefficient continuity assumptions

Complementary and particular solutions

Undetermined coefficients

Resonance adjustments

Variation of parameters

Variable coefficient forcing

Check yourself

How this lesson connects