Solution of linear differential equation

The differential equation of the form

$\displaystyle y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_{1}(x)y^{\prime} + a_{0}(x)y = f(x)$

is called the nth-order linear differential equation. $a_{i}(x)$ is called the coefficient function. $f(x)$

is called the input function.

If $f(x) \equiv 0$ , then the differntial equation is called the homogeneous equation.

We denote the left-hand side as $L(y)$ . Then the differential equation is expressed as

$\displaystyle L(y) = f(x)$

is called the differential operator and has linearlity. That is for any solutions $y_{1},y_{2}$ and constants $c_{1},c_{2}$ ,

$\displaystyle L(c_{1}y_{1} + c_{2}y_{2}) = c_{1}L(y_{1}) + c_{2}L(y_{2})$

Theorem 2..1 The solutions of homogeneous differential equations form a vector space.

We review the vector space.

A sum of two vectors A and B is expressed as A $+$ B and is equal to the diagonal of the parallelogram formed by A and B.

**Figure 2.1:** vector addition and scalar multiplication
$\begin{figure}\begin{center} \includegraphics[width=14cm]{DFQ/Fig0-1.eps} \end{center}\vspace{-1.6cm} \end{figure}$

1. A sum of two vetors is a vector (closure)
2. For any vectors A and B,A+B = B+A (commutative law)
3. For any vectors A,B,C, (A+B)+C = A+(B+C) (associative law)
4. Given any vector A, there exists a vector 0 satisying A+0 = A (existence of zero)
5. Given any vector A, there exists a vector B satisfying A+B = 0 (existence of inverse)
6. A scalar multiplication of a vector is a vector
7. For any real numbers $\alpha$ and $\beta$ , $\alpha$ ( $\beta$ A) = ( $\alpha\beta$ )A (associative law)
8. For any real numbers $\alpha$ and $\beta$ , ( $\alpha + \beta$ )A = $\alpha$ A + $\beta$ A and for any vectors A and B, $\alpha$ (A+B) = $\alpha$ A + $\alpha$ B (distributive law)
9. 1A = A; 0A = 0; $\alpha$ 0 = 0 (1 is multiplicative identity)

Let $C(a,b)$ be the set of continuous functions on the interval $(a,b)$ . Let $PC(a,b)$ be the set of piecewise continuous functions on $(a,b)$ , ^2.1

$C(a,b) = \{f(x) : f(x)$ is continuous on $(a,b)$ }, $PC(a,b) = \{f(x) : f(x)$ is piecewise continuous on $(a,b)$ }.

For $f(x)$ and $g(x)$ in $C(a,b)$ or $PC(a,b)$ , we define the addition on the scalar multiplication as follows:
1. $f+g$ is a function of $x$ whose value is equal to $f(x)+g(x)$ .
2. $\alpha f$ is a function of $x$ whose value is equal to $\alpha f(x)$ .

Example 2..1 Given $f(x) = x , g(x) = x^2$

, find $f+g, \frac{1}{2}f, 2g$ .

SOLUTION $(f+g)(x) = f(x) + g(x) = x + x^2$
$(\frac{1}{2}f)(x) = \frac{1}{2}f(x) = \frac{x}{2}$
$(2g)(x) = 2g(x) = 2x^2$ $\ \blacksquare$

Theorem 2..2 $C(a,b)$

with the operation defined above is a vector space. Then the function $f(x)$

belongs to $C(a,b)$

is called a vector.

Proof The solutions of a homogeneous differential equation are differentiable. Thus, the set of solutions are the subset of continuous functions. Then to show $C(a,b)$ is a vector space, it is enough to show the linear combination of the solutions $y_1$ and $y_2$ is a solution. Let $L(y_{1}) \equiv 0, \ L(y_{2}) \equiv 0, y_{3} = c_{1}y_{1} + c_{2}y_{2}$ . Then

$\displaystyle L(y_{3})$	$\displaystyle =$	$\displaystyle L(c_{1}y_{1} + c_{2}y_{2})$
	$\displaystyle =$	$\displaystyle L(c_{1}y_{1}) + L(c_{2}y_{2})$
	$\displaystyle =$	$\displaystyle c_{1}L(y_{1}) + c_{2}L(y_{2})$
	$\displaystyle =$	0

Thus

is again a solution. $\ \blacksquare$

The set of solutions of homogeneous equation becomes a vector space. So, we call this solution space.

Theorem 2..3 The basis of the solution space is a set of $n$

independent solutions of homogeneous differential equation.

The determinant of the following matrix is called Wronskian determinant. Let $y_1, y_2$ be the solutions of the differential equation. Then

$\displaystyle W(y_{1},y_{2},\ldots,y_{n}) = \left\vert\begin{array}{llll} y_{1}... ... y_{1}^{(n-1)},&y_{2}^{(n-1)},&\cdots,&y_{n}^{(n-1)} \end{array}\right \vert .$

Theorem 2..4 If $y_{1},y_{2},\ldots,y_{n}$ are the solutions of the homogeneous equation on the interval $[a,b]$

, then $W(y_{1},y_{2},\ldots,y_{n})$ is either 0 or never 0 0n the interval $[a,b]$

Proof For $n=2$ Since $y_{1}$ and $y_{2}$ are solutions of $L(y) = y^{\prime\prime} + a_{1}y^{\prime} + a_{0}y = 0$ , we have

$\displaystyle \frac{d}{dx}W(y_{1},y_{2})$	$\displaystyle =$	$\displaystyle \left\vert\begin{array}{rr} y_{1}^{\prime}&y_{2}^{\prime}\\ y_{1... ...y_{1}&y_{2}\\ y_{1}^{\prime\prime}&y_{2}^{\prime\prime} \end{array}\right\vert$
	$\displaystyle =$	$\displaystyle \left\vert\begin{array}{rr} y_{1}&y_{2}\\ -a_{1}y_{1}^{\prime}-a_{0}y_{1}&-a_{1}y_{2}^{\prime}-a_{0}y_{2}\end{array}\right\vert$
	$\displaystyle =$	$\displaystyle -a_{1}W(y_{1},y_{2}) .$

Thus

$\displaystyle \frac{d}{dx}W(y_{1},y_{2}) + a_{1}(x)W(y_{1},y_{2}) = 0.$

This is a linear differential equation. Thus the integrating factor $\mu(x) = \exp(\int_{x_{0}}^{x}a_{1}(t)dt)$ . Multiplying $\mu$ to get

$\displaystyle \frac{d}{dx}(\mu(x)W(y_{1},y_{2})) = 0$

Integrating

$\displaystyle W(y_{1},y_{2})(x) = c\exp(-\int_{x_{0}}^{x}a_{1}(t)dt) \ (c:$ constant $\displaystyle ) .$

Now let $x = x_{0}$ . Then $c = W(y_{1},y_{2})(x_{0})$ and

$\displaystyle W(y_{1},y_{2})(x) = W(y_{1},y_{2})(x_{0})\exp(-\int_{x_{0}}^{x}a_{1}(t)dt)$

This becomes 0 or not is obvious $\ \blacksquare$

Theorem 2..5 If $y_{1},y_{2},\ldots,y_{n}$ are solutions of the homogeneous differential equation on $[a,b]$

, then the following conditions are equivalent.
(1) $\{y_{1},y_{2},\ldots,y_{n}\}$ are linearly independent on the interval $[a,b]$

.
(2) $W( y_{1},\ldots,y_{n}) \neq 0, \ (x \in I)$

Proof For $n=2$ , let the linear combination of $y_{1}$ and $y_{2}$ be 0. Then

$\displaystyle c_{1}y_{1} + c_{2}y_{2} = 0 .$

Differentiate with respect to $x$

. Then

$\displaystyle c_{1}y^{\prime}_{1} + c_{2}y^{\prime}_{2} = 0$

Now using Cramer's rule to find $c_1$

and

$\displaystyle c_{1} = \frac{0}{W(y_{1},y_{2})}, \ c_{2} = \frac{0}{W(y_{1},y_{2})} .$

If $\{y_{1},y_{2}\}$ are linearly independent, then $c_{1} = c_{2} = 0$ . Thus, $W(y_{1},y_{2}) \neq 0$ . Conversely, if $W(y_{1},y_{2}) \neq 0$ , then $c_{1} = c_{2} = 0$ . Thus, $\{y_{1},y_{2}\}$ are linearly independent $\ \blacksquare$

Example 2..2 Find the general solution of $y^{(4)} = 0$ .

SOLUTION The dimension of the solution space is 3. $y_{1} = 1, y_{2} = x, y_{3} = x^{2}, y_4 = x^3$ are solutions of the differential equation. So, we need to show they are linearly independent.

$\displaystyle W(1,x,x^{2},x^3) = \left\vert\begin{array}{rrrr} 1&x&x^{2}&x^3\\ 0&1&2x&3x^2\\ 0&0&2&6x\\ 0&0&0&6 \end{array}\right\vert = 12.$

Thus, Wronskian is not 0 and therefore linearly independent. The general solution is

$\displaystyle y = c_{1} + c_{2}x + c_{3}x^{2} + c_4 x^3\ensuremath{\ \blacksquare}$

Example 2..3 Suppose that $y = e^{mx}$ is a solution of $L(y) = y^{\prime\prime} + 3y^{\prime} + 2y = 0$ . Then find the fundamental solution.

SOLUTION Since

$\displaystyle L(e^{mx})$	$\displaystyle =$	$\displaystyle m^{2}e^{mx} + 3me^{mx} + 2e^{mx}$
	$\displaystyle =$	$\displaystyle e^{mx}(m^{2} + 3m + 2)$
	$\displaystyle =$	$\displaystyle e^{mx}(m+1)(m+2),$

implies that $L(e^{mx}) \equiv 0$ .

$\displaystyle W(e^{-x},e^{-2x}) = \left\vert\begin{array}{rr} e^{-x}&e^{-2x}\\ -e^{-x}&-2e^{-2x} \end{array}\right\vert = -e^{-2x} \neq 0$

Thus, $\{e^{-x},e^{-2x}\}$ are fundamental solutions. $\ \blacksquare$

Theorem 2..6 Suppose $y_{p}$ is an particular solution of $n$

the-order linear differential equation $L(y) = f(x)$

and $y_{c}$ is the general solution of $L(y) = 0$

. Then $y(x) = y_{c}(x) + y_{p}(x)$ is the general solution of $L(y) = f(x)$

Proof By the assumption, $L(y_{p}) = f(x), L(y_{c}) = 0$ . Now by the linearlity of $L$ , we have

$\displaystyle L(y_{c} + y_{p}) = L(y_{c}) + L(y_{p}) = f(x).$

Thus $y_{c} + y_{p}$ is a solution of $L(y) = f(x)$

. Since $y_{c}(x)$ contains $n$

constants, $y_{c}(x) + y_{p}(x)$ is the general solution. $\ \blacksquare$

$y_{c}(x)$ is called the complementary solution. By this theorem, if an particular solution of $L(y) = f(x)$ is found, then to find the general solution, it is enough to find a complementary solution of $L(y) = 0$ .

Example 2..4 Show $x - 1$

is an particular solution of $y^{\prime\prime} + 3y^{\prime} + 2y = 2x + 1$ . Then find the general solution.

SOLUTION Since $y = x-1, y^{\prime} = 1, y^{\prime\prime} = 0$ , $y^{\prime\prime} + 3y^{\prime} + 2y = 2x + 1$ . Also, the complementary solution of $L(y) = 0$ is given in example2.2. Thus

$\displaystyle y_{c}(x) = c_{1}e^{-x} + c_{2}e^{-2x} .$

Therefore, the general solution is

$\displaystyle y(x) = \underbrace{c_{1}e^{-x} + c_{2}e^{-2x}}_{y_{c}} + \underbrace{x-1}_{y_{p}}\ensuremath{\ \blacksquare}$

If $f(x)$ is given as a sum of $f_{1}(x)$ and $f_{2}(x)$ , then it is better to consider $L(y) = f_{1}(x)$ and $\ L(y) = f_{2}(x)$ .

Theorem 2..7 Suppose that $y_{p_{1}}$ is a solution of $L(y) = f_{1}(x)$ , and $y_{p_{2}}$ is a solution of $\ L(y) = f_{2}(x)$ . Then $y_{p_{1}} + y_{p_{2}}$ is a solution of $L(y) = f_{1}(x) + f_{2}(x)$ .

Proof.

$\displaystyle L(y_{p_1} + y_{p_2}) = L(y_{p_1}) + L(y_{p_2}) = f_{1}(x) + f_{2}(x)$

For example, to find the general solution of the differential equation

$\displaystyle y^{\prime\prime} - 5y^{\prime} + 2y = \sin{x} + x^{2},$

find (1) $y_{c}$ of $y^{\prime\prime} - 5y^{\prime} + 2y = 0$
(2) $y_{p_{1}}$ of $y^{\prime\prime} - 5y^{\prime} + 2y = \sin{x}$
(3) $y_{p_{2}}$ of $y^{\prime\prime} - 5y^{\prime} + 2y = x^{2}$
and the linear combination

$\displaystyle y = y_{c} + y_{p_{1}} + y_{p_{2}}$

Subsections

Exercise