Chapter 36Second-Order Linear Nonhomogeneous Differential Equations and Euler's Equation

36.1 Structure of the General Solution of a Nonhomogeneous Equation and Variation of Parameters

First, consider the second-order linear nonhomogeneous differential equation with constant coefficients $a,b$

It is helpful to think of the right-hand side $R(x)$ as an “input” or an “external force” applied to the equation (if $R(x)=0$, it is the homogeneous equation we studied in Chapter 35). We write the homogeneous equation corresponding to ① as

As the theorem proved in Chapter 35 shows, the general solution $y$ of ① has the form

Here $Y$ is the general solution of ② (written with the fundamental solutions $y_1,y_2$ as $Y=C_1y_1+C_2y_2$), and we already learned how to find $Y$ in Chapter 35. On the other hand, $y_0$ need only be a single solution satisfying ①—we call this a particular solution of ①. The goal of this chapter is to master concrete methods for finding this $y_0$.

Let us investigate how to choose $C_1(x),C_2(x)$ so that ③ really satisfies ①. Differentiating both sides of ③ with respect to $x$ (using the product rule twice), we get

Here we use a little trick. We have not yet imposed a single condition on the two unknown functions $C_1(x),C_2(x)$ (③ says only that “$y_0$ has this form,” and $C_1,C_2$ themselves are free). So, to make the computation easier, we choose to impose the condition that the first parenthesized part be exactly $0$

(the computation that follows will confirm that, under this condition, $C_1,C_2$ can indeed be determined without contradiction). Then

so that $C_1',C_2'$ do not appear in the expression for $y_0'$. Differentiating ⑤ once more with respect to $x$, we obtain

Since $y_0$ is a solution of ①, substituting ③, ⑤, and ⑥ into the left-hand side of ① must give a result equal to $R(x)$. Carrying out the substitution, we get

Regrouping the right-hand side in terms of $C_1,C_2$, we obtain

The underbraced parts are $0$ because $y_1,y_2$ are both fundamental solutions of ② (the homogeneous equation). Therefore only $C_1'y_1'+C_2'y_2'$ remains on the left-hand side of ①, and this equals $R(x)$:

Regard ④ and ⑦ as a system of linear equations in the unknowns $C_1',C_2'$ and rewrite it in matrix form (using the notation for the product of a matrix and a vector that we learned in Chapters 8 and 9).

The determinant of this $2\times2$ matrix is exactly the Wronskian $W(y_1,y_2)=y_1y_2'-y_2y_1'$ of $y_1,y_2$. Since $y_1,y_2$ are linearly independent fundamental solutions of ②, $W(y_1,y_2)\neq0$ always holds, as we learned in Chapter 35. Because the determinant is not $0$, this matrix is invertible (it has an inverse matrix), and the formula for the inverse of a $2\times2$ matrix (Chapter 8) is

Multiplying both sides of ⑧ on the left by this inverse, we obtain

That is,

(The right-hand sides are functions of $x$, so integrating both sides with respect to $x$ gives $C_1(x),C_2(x)$ themselves.) Integrating, we obtain

Substituting these into ③, we can write down the particular solution $y_0$ explicitly.

Let us apply Eq. \eqref{eq:36-varparam} to a concrete equation.

36.2 Solving the Equation When Only One Fundamental Solution Is Known (Reduction of Order)

The method of variation of parameters in Section 36.1 is powerful, but it presupposed that both fundamental solutions of the homogeneous equation were known. What should we do, then, when we can find only one fundamental solution? In fact, even from a single fundamental solution there is a way to derive the general solution of the nonhomogeneous equation—that is, both a particular solution $y_0$ and another fundamental solution. This method is called reduction of order.

The coefficients $P(x),Q(x)$ may be constants or functions of $x$ (as we saw in the Caution in Section 36.1, the derivation does not use the fact that the coefficients are constants). For the nonhomogeneous equation

we write the corresponding homogeneous equation as

Suppose that only one fundamental solution of this homogeneous equation is known: $y_1(\neq0)$.

Differentiate both sides of ③ once and twice (product rule).

Since we assumed that ③ is a solution of ①, the result of substituting ③, ④, ⑤ into the left-hand side of ① equals $R(x)$. Actually carrying out the substitution, we get

Regroup the right-hand side by collecting the terms in $u,u',u''$.

The underbraced part is $0$ because $y_1$ is a solution of the homogeneous equation ($y_1''+P(x)y_1'+Q(x)y_1=0$). Hence only the terms involving the first and second derivatives of $u$ remain, and we obtain

Dividing both sides by $y_1$ ($\neq0$) turns this into an equation for $u$.

Now let $u'=p(x)$ (so that $u''=p'$), and abbreviate $A(x)=\dfrac{2y_1'}{y_1}+P(x)$ and $B(x)=\dfrac{R(x)}{y_1}$. Then we obtain

This is the form the equation takes.

Indeed, the solution formula

can be used as it stands. Once $p(x)$ is found, we integrate it once more with respect to $x$ to obtain $u(x)$, and finally substitute into ③, $y=u(x)y_1$; this completes the general solution of ①.

36.3 Euler's Differential Equation — Discovering the Trial Solution $x^\lambda$

So far we have mainly dealt with equations with constant coefficients (Sections 36.1 and 36.2). From here on, we turn to second-order linear differential equations whose coefficients are functions of $x$. The most fundamental of these, and the one with the widest range of applications, is the following Euler's differential equation.

36.3.1 Why try $x^\lambda$? — Examining a simple case

Recall that in Chapter 35, when we substituted the exponential function $y=e^{\lambda x}$ into the constant-coefficient homogeneous equation $y''+ay'+by=0$, a quadratic equation in $\lambda$ (the characteristic equation) appeared. For Euler's equation as well, we begin in the same way, by finding the “form” of the solutions. To keep things simple, consider the case $b=0$, $$ x^2\frac{\dd y'}{\dd x}+ax\cdot y'=0 $$ (noting that $y''=\dfrac{\dd y'}{\dd x}$, we regard this as a first-order equation for $y'$). For $x\neq0$, dividing both sides by $x\cdot y'$ turns it into a separable equation.

Integrating both sides (by the method of separation of variables from Chapter 33) gives

Let $C_1=\ln|C_2|$ (where $C_2\neq0$ is a constant) and combine the right-hand side into a single logarithm; this gives $\ln|y'|=\ln\left|C_2x^{-a}\right|$, and exponentiating both sides yields

Here $C_2$ is an arbitrary constant, which may be positive or negative. We integrate this once more with respect to $x$. For $a\neq1$, using the integration formula for power functions $\displaystyle\int x^{-a}\dd x=\frac{x^{1-a}}{1-a}$ (since $-a\neq-1$), we get

Regrouping the constants by setting $C_0=\dfrac{C_2}{1-a}$, we obtain

Thus $x^{1-a}$ (together with the constant function $x^0=1$) emerges as a candidate for a fundamental solution. In other words, we are led to the conjecture that, in the general case as well, the fundamental solutions have the form $x^\lambda$ ($\lambda$ some constant).

36.3.2 The general case: substituting $x^\lambda$ directly

Based on this conjecture, let us actually substitute $y=x^\lambda$ ($\lambda$ a constant) into Euler's equation in its general form $x^2y''+axy'+by=0$. Since $y'=\lambda x^{\lambda-1}$ and $y''=\lambda(\lambda-1)x^{\lambda-2}$, we have

(note that $x^\lambda$ is a common factor of all three terms—this is the reason why $x^\lambda$ fits Euler's equation so well). Factoring out $x^\lambda$, we obtain

The underlined part is called the characteristic equation. Since $x\neq0$ implies $x^\lambda\neq0$, for the left-hand side to be $0$ the underlined part itself must be $0$. Expanding and collecting terms in $\lambda$, we get

which is a quadratic equation in $\lambda$.

36.4 Solving Euler's Differential Equation — Real Roots, Repeated Roots, and Complex Roots

In Section 36.3 we derived the characteristic equation $\lambda^2+(a-1)\lambda+b=0$ of Euler's equation $x^2y''+axy'+by=0$. Depending on the type of roots of this quadratic equation (the sign of the discriminant), the form of the general solution splits into three cases. This has exactly the same logical structure as the three-way classification of constant-coefficient homogeneous equations that we learned in Chapter 35—in fact, with the substitution $x=e^t$, which we will study in Chapter 37, Euler's equation can be transformed completely into an equation with constant coefficients. Here we examine the three cases directly, within the framework of Euler's equation, without using that substitution.

36.4.1 The case of two distinct real roots

When the roots $\lambda_1,\lambda_2$ ($\lambda_1\neq\lambda_2$) of the characteristic equation are real, the fundamental solutions are $x^{\lambda_1},x^{\lambda_2}$, as we already saw in Section 36.3. Just to be safe, let us confirm their linear independence with the Wronskian.

Since $\lambda_1\neq\lambda_2$ and $x\neq0$, we have $W(x^{\lambda_1},x^{\lambda_2})\neq0$. Therefore $x^{\lambda_1},x^{\lambda_2}$ are linearly independent fundamental solutions, and the general solution is

.

36.4.2 The case of a repeated root

The characteristic equation $\lambda^2+(a-1)\lambda+b=0$ has a repeated root $\lambda_0$ when the discriminant is $0$, that is, when $(a-1)^2-4b=0$.

In this case one fundamental solution is $x^{\lambda_0}$, but a second-order differential equation needs two fundamental solutions (Chapter 35: the solution space is two-dimensional). So we look for another fundamental solution by setting $y=u(x)\cdot x^{\lambda_0}$ (the same idea as the reduction of order in Section 36.2).

Substituting these into $x^2y''+axy'+by$, we obtain

The expression in the braces becomes $0$ because $\lambda_0$ is a root of the characteristic equation. Hence the condition that this equal $0$ (the homogeneous equation) becomes

Here, since $\lambda_0=-\dfrac{a-1}{2}$, that is, $2\lambda_0=-(a-1)=1-a$, we have

This is a convenient relation (an identity that always holds in the case of a repeated root). Since $x^{\lambda_0+1}\neq0$ ($x\neq0$), the expression in the braces must be $0$, which leaves the following equation for $u$:

Setting $v=u'$ gives $xv'+v=0$, that is, the separable equation

Integrating (since we take $x\gt0$, no absolute value is needed), we obtain

Hence $u=\displaystyle\int v\,\dd x=-C_1\ln x+C_2$ (the way the sign is attached depends on how the arbitrary constants are chosen, so from here on we may rewrite $C_1\to-C_1$ and write $u=C_1\ln x+C_2$). Since it is $u(x)$ itself that we want to use for the fundamental solution, we choose $C_1=1,C_2=0$ and take $u(x)=\ln x$, which gives another fundamental solution

Checking the Wronskian, we find

so the linear independence is confirmed.

36.4.3 The case of complex roots

Consider the case where the characteristic equation has complex roots $\lambda_1=\alpha+i\beta,\ \lambda_2=\alpha-i\beta$ ($\alpha,\beta\in\R$, $\beta\neq0$, and $i$ is the imaginary unit). In this case too, the fundamental solutions are formally $x^{\alpha+i\beta},x^{\alpha-i\beta}$, and the general solution can be written as

but we have not yet explained the meaning of $x^{i\beta}$, a “real number raised to a complex power.”

Using Euler's formula, we get

(Here $e^{-i\beta\ln x}=\cos(\beta\ln x)-i\sin(\beta\ln x)$ follows from Euler's formula with $\theta\to-\theta$, using the facts that $\cos$ is an even function and $\sin$ is an odd function.) Collecting the terms in $\cos(\beta\ln x)$ and in $\sin(\beta\ln x)$ separately, we obtain

Now let $C_1=C_1'+C_2',\ C_2=i(C_1'-C_2')$ (if $C_1',C_2'$ are chosen suitably from the complex numbers, $C_1,C_2$ become arbitrary real constants). Then

We thus obtain a real-valued general solution that does not contain the imaginary unit $i$. Checking the Wronskian, we find

.

36.5 General Second-Order Linear Differential Equations — Conditions for Finding Fundamental Solutions

Euler's equation was a special example that can be solved even though its coefficients are functions of $x$. Here we widen our view further and consider the most general second-order linear differential equation, in which the coefficients are arbitrary functions $P(x),Q(x)$:

Even if the coefficients are not constant, as long as we know the fundamental solutions $y_1,y_2$ (solutions of the homogeneous equation) and a particular solution $y_0$, the general solution takes the form

exactly as in the discussions of Sections 36.1 and 36.2. Moreover, Formula 36.1 (variation of parameters) for finding a particular solution $y_0$ can also be used as it is even when the coefficients are not constant, as we confirmed in the Caution in Section 36.1:

The problem is how to find the two fundamental solutions $y_1,y_2$ themselves. For general $P(x),Q(x)$ there is no universal method for finding fundamental solutions. However, only when $P(x),Q(x)$ satisfy certain special conditions can we sometimes find a fundamental solution of the form $x^\lambda$ or $e^{\lambda x}$ directly, simply by substitution.

36.5.1 Condition (I): When $x^\lambda$ is a fundamental solution

Let us substitute $y=x^\lambda$ ($\lambda$ is a constant) into the left-hand side $\dfrac{\dd^2y}{\dd x^2}+P(x)\dfrac{\dd y}{\dd x}+Q(x)y$ of the general equation. Since $y'=\lambda x^{\lambda-1}$ and $y''=\lambda(\lambda-1)x^{\lambda-2}$, we have

Factoring $x^{\lambda-2}$ ($\neq0$) out of the right-hand side, we obtain

Therefore, if we can choose the constant $\lambda$ cleverly so that

holds, then the contents of the curly braces become $0$, and we see that $y=x^\lambda$ is one of the fundamental solutions of this equation.

36.5.2 Condition (II): When $e^{\lambda x}$ is a fundamental solution

Similarly, substituting $y=e^{\lambda x}$ and using $y'=\lambda e^{\lambda x}$ and $y''=\lambda^2e^{\lambda x}$, we have

Thus, if we can choose the constant $\lambda$ cleverly so that

holds, then $y=e^{\lambda x}$ is one of the fundamental solutions.

36.6 Reduction to Normal Form

Conditions (I) and (II) of Section 36.5 can be used only when $P(x),Q(x)$ have rather special forms. For general $P(x),Q(x)$, we cannot hope to find fundamental solutions directly. However, we can always rewrite the equation itself in a form that is easier to work with. Here we learn a transformation that eliminates the first-derivative term $y'$.

Differentiating both sides of $y=uv$ once and twice (by the product rule), we obtain

Substituting these into ① gives

Regrouping this according to the order of the derivative of $u$ ($u,u',u''$), we obtain

Now recall that we have not yet specified $v(x)$ at all. If we deliberately choose the coefficient $2v'+P(x)v$ of $u'$ ourselves so that

then the $u'$ term vanishes from ②, and only the $u''$ term and the $u$ term remain—this is the mechanism for producing the “normal form.” Since this condition is a first-order separable equation (Chapter 33),

(Since $v$ is a function of which “we need to find just one,” no integration constant is needed; we may take $C=0$.) For the $v$ chosen in this way, we have $v'=-\dfrac12P(x)v$, and differentiating this once more gives $v''=-\dfrac12\big(P'(x)v+P(x)v'\big)$. Substituting these into the coefficient $v''+P(x)v'+Q(x)v$ of $u$ in ②, we get

Substituting $v'=-\dfrac12Pv$ here as well gives $\dfrac12Pv'=\dfrac12P\cdot\left(-\dfrac12Pv\right)=-\dfrac14P^2v$, so we obtain

Hence ② becomes

and dividing both sides by $v(\neq0)$ gives

If we set $I(x)=Q(x)-\dfrac12P'(x)-\dfrac14P^2(x)$ and $J(x)=\dfrac{R(x)}{v(x)}$, this becomes

a form that has no first-derivative term.

36.7 Phase Space and Differential Equations of Simple-Harmonic Type

To close this chapter, we take another look at the differential equation that appears most often in physics—the equation of simple harmonic motion—from a geometric point of view different from the ones we have used so far. When a weight of mass $m$ is attached to a spring with spring constant $k$ and moves, its position $x(t)$ ($t$ is time) satisfies the following equation, which follows from Newton's equation of motion ($F=ma$, with Hooke's law $F=-kx$):

In what follows, we use a notation common in physics: we write the first derivative with respect to time $t$ as $\dot x=\dfrac{\dd x}{\dd t}$ and the second derivative as $\ddot x=\dfrac{\dd^2x}{\dd t^2}$, putting dots over $x$ (Newton's notation). This means exactly the same thing as the $y',y''$ we have used so far; it is the form preferred when one wants to emphasize that the independent variable is $t$. In this notation, the equation of motion above becomes $m\ddot x+kx=0$, which is precisely the homogeneous equation with constant coefficients of Chapter 35; and the fact that the characteristic equation $m\lambda^2+k=0$ gives $\lambda=\pm i\sqrt{k/m}$, and hence trigonometric solutions, can also be confirmed by the method we already know. Here, however, we deliberately do not use the characteristic equation, and instead try to solve the equation directly from the idea of energy conservation.

36.7.1 The energy integral

Multiply both sides of the equation by $v\Big(=\dfrac{\dd x}{\dd t}\Big)$ and integrate with respect to $t$ (this operation is called the energy integral).

By the relation $\dfrac{\dd}{\dd t}\left(\dfrac12mv^2\right)=mv\dfrac{\dd v}{\dd t}$, the left-hand side becomes the integral that produces $\dfrac12mv^2$, and likewise the right-hand side becomes the integral that produces $\dfrac12kx^2$, so we obtain

The first term on the left is the kinetic energy and the second term is the potential energy of the spring, so this equation is precisely the law of conservation of mechanical energy. Take the initial conditions $x(0)=A$ (we first stretch the spring by $A$ and release it) and $\dot x(0)=0$ (initial velocity $0$). Then at $t=0$ we have $C=\frac12m\cdot0^2+\frac12kA^2=\frac12kA^2$, so the equation becomes

Dividing both sides by $\frac12k$, we obtain

Now let $\omega=\sqrt{k/m}$ (the angular frequency). Since $\dfrac{m}{k}=\dfrac1{\omega^2}$, if we give the new name $V$ to the quantity $\sqrt{m/k}\,\dot x=\dfrac{\dot x}{\omega}$, we obtain

The familiar equation of a circle appears.

36.7.2 The circle in phase space and the angle $\theta$

Eq. Ⓐ represents a circle of radius $A$ centered at the origin in the plane whose horizontal axis is $x$ and whose vertical axis is $V$ (this $(x,V)$-plane is called the phase space). A point on the circle can be parametrized, using the angle $\theta$ (the angle measured from the $x$-axis), as follows:

(Since $\cos^2\theta+\sin^2\theta=1$, this indeed satisfies Eq. Ⓐ.)

From ① and ②, we compute $\dot x$ in two ways. First, differentiating ① directly with respect to $t$ (note that $\theta$ is also a function of $t$, so this is the differentiation of a composite function), we obtain

On the other hand, combining $V=\dot x/\omega$ (the definition in Ⓐ) with ② gives

Since we have written the same $\dot x$ in two ways, the two expressions are equal:

(As long as $\sin\theta\neq0$, we may divide both sides by $A\sin\theta$. The instants at which $\sin\theta=0$ are isolated points, so by continuity the following conclusion holds for all $t$.)

This is a very simple first-order differential equation for $\theta$. Integrating it gives

Substituting this into ① gives $x(t)$.

(We used the fact that $\cos$ is an even function, $\cos(-\phi)=\cos\phi$.)

Related simulator: Damped and Forced Oscillations (the simple harmonic motion of this section is the special case with no damping and no external force; by bringing the angular frequency of the external force close to the natural frequency, you can check numerically how the resonance seen in Exercise 36.1 occurs).

36.8 Summary and Exercises

36.8.1 Summary

• The general solution of the second-order linear nonhomogeneous equation $y''+ay'+by=R(x)$ has the form $y=y_0+Y$ ($Y$: the general solution of the corresponding homogeneous equation; $y_0$: any one particular solution satisfying ①).
• Variation of parameters (Section 36.1): When both fundamental solutions $y_1,y_2$ of the homogeneous equation are known, set $y_0=C_1(x)y_1+C_2(x)y_2$ and impose the condition $C_1'y_1+C_2'y_2=0$ of our own accord; this reduces the problem of finding $C_1',C_2'$ to a $2\times2$ system of linear equations (a matrix and its inverse) (Formula 36.1). This is a general formula that holds even when the coefficients are functions of $x$.
• Reduction of order (Section 36.2): Even when only $y_1$ is known as a fundamental solution, setting $y=u(x)y_1$ reduces the problem to a first-order linear differential equation for $u$, and the general solution of the nonhomogeneous equation (both a particular solution and the other fundamental solution) then comes out one after another (Theorem 36.1).
• Euler's differential equation $x^2y''+axy'+by=0$ (Sections 36.3 and 36.4): Substituting the trial solution $x^\lambda$ produces the characteristic equation $\lambda^2+(a-1)\lambda+b=0$ (Formula 36.2). The general solution falls into three cases according to the type of roots: for distinct real roots, $y=C_1x^{\lambda_1}+C_2x^{\lambda_2}$; for a repeated root, $y=C_1x^{\lambda_0}+C_2x^{\lambda_0}\ln x$ (Theorem 36.2); for complex roots, $y=C_1x^\alpha\cos(\beta\ln x)+C_2x^\alpha\sin(\beta\ln x)$ (Theorem 36.3).
• Even for a general variable-coefficient equation $y''+P(x)y'+Q(x)y=0$, $x^\lambda$ is directly obtained as one of the fundamental solutions if condition (I) $\lambda(\lambda-1)+\lambda xP(x)+x^2Q(x)=0$ is satisfied, and $e^{\lambda x}$ is directly obtained as one if condition (II) $\lambda^2+\lambda P(x)+Q(x)=0$ is satisfied (Theorem 36.4, Section 36.5).
• Reduction to normal form (Section 36.6): Setting $y=u(x)v(x)$ with $v(x)=e^{-\frac12\int P(x)\dd x}$ gives the normal form $u''+I(x)u=J(x)$ ($I=Q-\frac12P'-\frac14P^2$), in which the first-derivative term has disappeared (Theorem 36.5).
• Phase space (Section 36.7): The differential equation of simple-harmonic type, $m\ddot x+kx=0$, can be solved geometrically as a uniform rotation on the circle $x^2+V^2=A^2$ in the $(x,V)$ plane ($V=\dot x/\omega$), and the solution is $x(t)=A\cos(\omega t-\theta_0)$ (Theorem 36.6). “Oscillation” and “rotation” are two views of the same phenomenon.

36.8.2 Exercises

36.8.3 References