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4.1 Preliminary Theory–Linear Equations: Class Notes

4.1a

Definition

The equations

$${\displaystyle a_n(x)\frac{d^{n}y}{d{x}^n}+a_{n-1}(x)\frac{d^{n-1}y}{d{x}^{n-1}}+\cdots +a_1(x)\frac{dy}{dx}+a_0(x)y=0(1)}$$

and

$${\displaystyle a_n(x)\frac{d^{n}y}{d{x}^n}+a_{n-1}(x)\frac{d^{n-1}y}{d{x}^{n-1}}+\cdots +a_1(x)\frac{dy}{dx}+a_0(x)y=g(x)(2)}$$

are $n$th-order, linear differential equations. (1) is homogeneous, and if $g(x)\not\equiv 0$, then (2) is nonhomogeneous.

Abbreviations: hom, homog for homogeneous.

Definition

The points $y(x_0)=y_0,y\prime (x_0)=y_1,\dots ,y^{(n-1)}(x_0)=y_{n-1}$ are initial conditions. An initial-value problem (IVP) is (1) or (2), together with initial conditions. In the event that we have information such as $y(a)=y_0,y(b)=y_1$, then we have a boundary-value problem (BVP).

In an initial-value problem, every condition is given at the same point $x_0$. In a boundary-value problem, the conditions are given at two different points $a$ and $b$. The conditions can also involve derivatives, for example

$$y(a)=y_0,y\prime (b)=y_1.$$

Theorem: Existence of a Unique Solution

Let $a_n(x),a_{n-1}(x),\dots ,a_1(x)$ and $g(x)$ be continuous on an interval $I$ and let $a_n(x)\ne 0$ for every $x$ in this interval. If $x=x_0$ is any point in this interval, then a solution $y(x)$ of the initial-value problem (2) exists on the interval and is unique.

These two important criteria will be assumed henceforth in the statement of theorems about linear DEs.

A BVP can have zero, one, or infinitely many solutions.

Example

The given two-parameter family is a solution of the indicated differential equation on the interval $(-\infty ,\infty )$. Determine whether a member of the family can be found that satisfies the boundary conditions.

$$y=c_1{x}^2+c_2{x}^4+3;x^{2}y″-5xy\prime +8y=24$$

(a) $y(-1)=0,y(1)=4$  (b) $y(0)=1,y(1)=2$

(c) $y(0)=3,y(1)=0$  (d) $y(1)=3,y(2)=15$

Solution

Substitute each boundary condition into $y=c_1{x}^2+c_2{x}^4+3$ and see whether the resulting equations in $c_1$ and $c_2$ can be solved.

(a)

$$\begin{array}{cc}\hfill {\displaystyle y(-1)=0}& {\displaystyle \Rightarrow c_1+c_2+3=0\Rightarrow c_1+c_2=-3}\hfill \\ \hfill {\displaystyle y(1)=4}& {\displaystyle \Rightarrow c_1+c_2+3=4\Rightarrow c_1+c_2=1}\hfill \end{array}$$

The two equations contradict each other ($c_1+c_2$ cannot equal both $-3$ and $1$).

(b)

$$y(0)=1\Rightarrow c_1(0)+c_2(0)+3=1\Rightarrow 3=1$$

This is false for every choice of $c_1$ and $c_2$.

(c)

$$\begin{array}{cc}\hfill {\displaystyle y(0)=3}& {\displaystyle \Rightarrow 3=3\text{(always true)}}\hfill \\ \hfill {\displaystyle y(1)=0}& {\displaystyle \Rightarrow c_1+c_2+3=0\Rightarrow c_2=-3-c_1}\hfill \end{array}$$

Only one condition restricts the constants, so $c_1$ can be any real number:

(d)

$$\begin{array}{cc}\hfill {\displaystyle y(1)=3}& {\displaystyle \Rightarrow c_1+c_2+3=3}\hfill \\ \hfill {\displaystyle y(2)=15}& {\displaystyle \Rightarrow 4c_1+16c_2+3=15}\hfill \end{array}$$

The first equation gives $c_1=-c_2$; then the second becomes $12c_2=12$, so

$$c_1=-1,c_2=1.$$

Tip: The four parts show all three possibilities for a BVP: (a) and (b) have no solution, (c) has infinitely many, and (d) has exactly one.

4.1b

Definition

A set of functions $f_1(x),f_2(x),\dots ,f_n(x)$ is said to be linearly dependent on an interval $I$ if there exist constants $c_1,c_2,\dots ,c_n$, not all zero, such that $c_1{f}_1(x)+c_2{f}_2(x)+\cdots +c_n{f}_n(x)=0$ for every $x$ in the interval. If the set of functions is not linearly dependent on the interval, it is said to be linearly independent.

Independent: $f_1(x)=e^x,f_2(x)=x^2+3,f_3(x)=\sin x$. The only way to have

$$c_1{e}^x+c_2(x^2+3)+c_3\sin x=0$$

for every $x$ is $c_1=c_2=c_3=0$.

Dependent: $y_1=3e^{2x},y_2=2e^{2x}$. These constants are not all zero:

$$-2\left(3e^{2x}\right)+3\left(2e^{2x}\right)=0.$$

Definition

Suppose each of the functions $f_1(x),f_2(x),\dots ,f_n(x)$ possesses at least $n-1$ derivatives. The determinant

$$W(f_1,f_2,\dots ,f_n)=\left|\begin{array}{cccc}{f}_1& f_2& \cdots & f_n\\ f_1\prime & f_2\prime & \cdots & f_n\prime \\ ⋮& ⋮& & ⋮\\ f_1^{(n-1)}& f_2^{(n-1)}& \cdots & f_n^{(n-1)}\end{array}\right|,$$

where the primes denote derivatives, is called the Wronskian of the functions.

Theorem: Criterion for Linearly Independent Solutions

Let $y_1,y_2,\dots ,y_n$ be $n$ solutions of the homogeneous $n$th-order differential equation (1) on an interval $I$. Then the set of solutions is linearly independent on $I$ if and only if $W(y_1,y_2,\dots ,y_n)\ne 0$ for every $x$ in the interval.

Note: The Wronskian of these solutions is either always zero or never zero on an interval.

Example

Determine whether the given set of functions is linearly independent on the interval $(-\infty ,\infty )$.

$f_1(x)=\cos 2x,f_2(x)=1,f_3(x)={\cos }^{2}x$

Solution

Each row of the Wronskian is the derivative of the row above it. For $f_3={\cos }^{2}x$, the chain rule gives $f_3\prime =-2\cos x\sin x$, which we rewrite with the double-angle identity as $-\sin 2x$; then $f_3″=-2\cos 2x$.

$$W(f_1,f_2,f_3)=\left|\begin{array}{ccc}\cos 2x& 1& {\cos }^{2}x\\ -2\sin 2x& 0& -\sin 2x\\ -4\cos 2x& 0& -2\cos 2x\end{array}\right|$$

The second column has two zeros, so expand down the second column. Its only nonzero entry, $1$, sits in a − position:

$$\begin{array}{cc}\hfill {\displaystyle W}& {\displaystyle =-1\left|\begin{array}{cc}-2\sin 2x& -\sin 2x\\ -4\cos 2x& -2\cos 2x\end{array}\right|}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-1(4\cos 2x\sin 2x-4\cos 2x\sin 2x)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =0}\hfill \end{array}$$

The Wronskian is zero, and in fact we can write down a dependence relation. Using the identity $\cos 2x=2{\cos }^{2}x-1$,

$$\cos 2x-2{\cos }^{2}x+1=0,\text{that is,}{f}_1+f_2-2f_3=0.$$

Theorem: Superposition Principle–Homogeneous Equations

Let $y_1,y_2,\dots ,y_k$ be solutions of the homogeneous $n$th-order differential equation (1) on an interval $I$. Then the linear combination $c_1{y}_1(x)+c_2{y}_2(x)+\cdots +c_k{y}_k(x)$, where the $c_i$, $i=1,2,\dots ,k$ are arbitrary constants, is also a solution on the interval.

For example, $y_1=e^{3x}$ and $y_2=e^{-5x}$ are solutions to $y″+2y\prime -15y=0$. Check $y=4e^{3x}$:

$$\begin{array}{cc}\hfill {\displaystyle y=4e^{3x}\Rightarrow y\prime }& {\displaystyle =4·3e^{3x}\Rightarrow y″=4·9e^{3x}}\hfill \\ \hfill {\displaystyle y″+2y\prime -15y}& {\displaystyle =4(9e^{3x}+2·3e^{3x}-15e^{3x})=4·0=0}\hfill \end{array}$$

The constant $4$ factors out, so $y_1=c_1{e}^{3x}$ is a solution for every $c_1\in ℝ$, and likewise $c_2{e}^{-5x}$. Then

$$y=c_1{e}^{3x}+c_2{e}^{-5x}$$

is also a solution for any $c_1,c_2\in ℝ$.

Definition

Any set $y_1,y_2,\dots ,y_n$ of $n$ linearly independent solutions of the homogeneous $n$th-order differential equation (1) on an interval $I$ is said to be a fundamental set of solutions on the interval.

The number of solutions matches the order of the equation.

Theorem: General Solution–Homogeneous Equations

Let $y_1,y_2,\dots ,y_n$ be a fundamental set of solutions of the homogeneous linear $n$th-order differential equation (1) on an interval $I$. Then the general solution of the equation on the interval is $y=c_1{y}_1(x)+c_2{y}_2(x)+\cdots +c_n{y}_n(x)$.

4.1c

Example

Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution.

$$x^{2}y″+xy\prime +y=0;\cos (\ln x),\sin (\ln x),(0,\infty )$$

Solution

Two things to show: (1) each function satisfies the equation (plug in and show the equation is true), and (2) the functions are linearly independent, $W\ne 0$.

Step 1: solutions. Let $y_1=\cos (\ln x)$. By the chain rule,

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y_1}& {\displaystyle =\cos (\ln x)}\hfill \\ \hfill {\displaystyle y_1\prime }& {\displaystyle =-\frac{\sin (\ln x)}{x}}\hfill \\ \hfill {\displaystyle y_1″}& {\displaystyle =\frac{\sin (\ln x)-\cos (\ln x)}{x^2}}\hfill \end{array}}$$

Substitute into the left side; the powers of $x$ cancel:

$$\begin{array}{cc}\hfill {\displaystyle x^2{y}_1″+x{y}_1\prime +y_1}& {\displaystyle =\sin (\ln x)-\cos (\ln x)+(-\sin (\ln x))+\cos (\ln x)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =0.}\hfill \end{array}$$

So $y_1$ is a solution. A similar computation shows $y_2=\sin (\ln x)$ is also a solution.

Step 2: linearly independent? Compute the Wronskian (main diagonal minus other diagonal):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle W}& {\displaystyle =\left|\begin{array}{cc}\cos (\ln x)& \sin (\ln x)\\ -\frac{\sin (\ln x)}{x}& \frac{\cos (\ln x)}{x}\end{array}\right|}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{{\cos }^2(\ln x)}{x}+\frac{{\sin }^2(\ln x)}{x}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{x}\ne 0\text{on }(0,\infty ).}\hfill \end{array}}$$

Yes: the functions are linearly independent, so they form a fundamental set of solutions.

Theorem: Existence of a Fundamental Set

There exists a fundamental set of solutions for the homogeneous linear $n$th-order differential equation (1) on an interval $I$.

Definition

Any function $y_p$, free of arbitrary parameters, that satisfies (2) is said to be a particular solution of the equation.

Theorem: General Solution–Nonhomogeneous Equations

Let $y_p$ be any particular solution of the nonhomogeneous linear $n$th-order differential equation (2) on an interval $I$ and let $y_1,y_2,\dots ,y_n$ be a fundamental set of solutions of the associated homogeneous differential equation (1) on $I$. Then the general solution of the equation on the interval is

$$y=c_1{y}_1(x)+c_2{y}_2(x)+\cdots +c_n{y}_n(x)+y_p(x),$$

where the $c_i$, $i=1,2,\dots ,n$ are arbitrary constants.

The first part is the general solution of the homogeneous DE (1), and the last part is a particular solution of the nonhomogeneous DE (2).

Theorem: Superposition Principle–Nonhomogeneous Equations

Let $y_{p_1},y_{p_2},\dots ,y_{p_k}$ be $k$ particular solutions of the nonhomogeneous linear $n$th-order differential equation (2) on an interval $I$ corresponding, in turn, to $k$ distinct functions $g_1,g_2,\dots ,g_k$. Then $y_p(x)=y_{p_1}(x)+y_{p_2}(x)+\cdots +y_{p_k}(x)$ is a particular solution of

$$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots +a_1(x)y\prime +a_0(x)y=g_1(x)+g_2(x)+\cdots +g_k(x).$$

In other words: find a particular solution for each piece of $g(x)$, then add them together.

Example

(a) Verify that $y_{p_1}=3e^{2x}$ and $y_{p_2}=x^2+3x$, respectively, are particular solutions of

$$y″-6y\prime +5y=-9e^{2x}\text{and}y″-6y\prime +5y=5x^2+3x-16.$$

(b) Use part (a) to find the particular solution of

$$y″-6y\prime +5y=5x^2+3x-16-9e^{2x}.$$

Solution

(a) For $y_{p_1}=3e^{2x}$: $y_{p_1}\prime =6e^{2x}$ and $y_{p_1}″=12e^{2x}$, so

$$12e^{2x}-6\left(6e^{2x}\right)+5\left(3e^{2x}\right)=(12-36+15)e^{2x}=-9e^{2x}.✓$$

For $y_{p_2}=x^2+3x$: $y_{p_2}\prime =2x+3$ and $y_{p_2}″=2$, so

$$\begin{array}{cc}\hfill {\displaystyle 2-6(2x+3)+5(x^2+3x)}& {\displaystyle =2-12x-18+5x^2+15x}\hfill \\ \hfill {\displaystyle }& {\displaystyle =5x^2+3x-16.✓}\hfill \end{array}$$

(b) The right side is the sum of the two right sides in part (a), so by the superposition principle the particular solution is $y_p=y_{p_1}+y_{p_2}$:

Going one step further: the associated homogeneous equation $y″-6y\prime +5y=0$ has general solution

$$y_c=c_1{e}^x+c_2{e}^{5x},$$

called the complementary solution. So the general solution of the nonhomogeneous equation is $y=y_c+y_p$: