Linear Equations

Some times we might think of a subset or subclass of differential equations which admit explicit solutions. This question is pertinent when we say that there are no means to find the explicit solution of $$\frac{dy}{dx} = f(x,y)$$ where $f$ is an arbitrary continuous function in $(x,y)$ in suitable domain of definition. In this context, we have a class of equations, called Linear Equations (to be defined shortly) which admit explicit solutions.

DEFINITION 7.4.1 (Linear/Nonlinear Equations)   Let $p(x)$ and $q(x)$ be real-valued piecewise continuous functions defined on interval $I = [a, b].$ The equation

$$y^\prime + p(x) y = q(x), \; x \in I$$ (7.4.1)

is called a linear equation, where $y^\prime$ stands for $$\frac{dy}{dx}.$$ The Equation (7.4.1) is called Linear non-homogeneous if $q(x) \neq 0$ and is called Linear homogeneous if $q(x) = 0$ on $I.$

A first order equation is called a non-linear equation (in the independent variable) if it is neither a linear homogeneous nor a non-homogeneous linear equation.

EXAMPLE 7.4.2  

1. The equation $y^\prime = \sin y$ is a non-linear equation.
2. The equation $y^\prime + y = \sin x$ is a linear non-homogeneous equation.
3. The equation $y^\prime + x^2 y = 0$ is a linear homogeneous equation.

Define the indefinite integral $P(x) = \int p(x) dx$ ( or $\int\limits_{a}^x p(s) ds$ ). Multiplying (7.4.1) by $e^{P(x)},$ we get

$$e^{P(x)} y^\prime + e^{P(x)} p(x) y = e^{P(x)} q(x) \;\; {\mbox{ or equivalently}} \;\; \frac{d}{dx} (e^{P(x)} y) = e^{P(x)} q(x).$$

On integration, we get

$$e^{P(x)} y = c + \int e^{P(x)} q(x) dx.$$

In other words,

$$y = c e^{-P(x)} + e^{-P(x)} \int e^{P(x)} q(x) dx$$ (7.4.2)

where $c$ is an arbitrary constant is the general solution of (7.4.1).

Remark 7.4.3   If we let $P(x) = \int\limits_{a}^x p(s) ds$ in the above discussion, (7.4.2) also represents

$$y = y(a) e^{-P(x)} + e^{-P(x)} \int\limits_a^x e^{P(s)} q(s) ds.$$ (7.4.3)

As a simple consequence, we have the following proposition.

PROPOSITION 7.4.4   $y = c e^{-P(x)}$ (where $c$ is any constant) is the general solution of the linear homogeneous equation

$$y^\prime + p(x) y = 0.$$ (7.4.4)

In particular, when $p(x) = k,$ is a constant, the general solution is $y = c e^{- k x},$ with $c$ an arbitrary constant.

EXAMPLE 7.4.5  

1. Comparing the equation $y^\prime = y$ with (7.4.1), we have
   $$p(x) = -1 \;\; {\mbox{ and }}\; \; q(x) = 0.$$
   Hence, $P(x) = \int (-1) dx = -x.$ Substituting for $P(x)$ in (7.4.2), we get $y = c e^x$ as the required general solution.

   We can just use the second part of the above proposition to get the above result, as $k = -1.$

2. The general solution of $x y^\prime = - y, \; x \in I \; (0 \not\in I)$ is $y = c x^{-1},$ where $c$ is an arbitrary constant. Notice that no non-zero solution exists if we insist on the condition $\lim\limits_{x \rightarrow 0, x> 0} y = 0.$

A class of nonlinear Equations (7.4.1) (named after Bernoulli $(1654 - 1705)$ ) can be reduced to linear equation. These equations are of the type

$$y^\prime + p(x) y = q(x) y^a.$$ (7.4.5)

If $a = 0$ or $a = 1,$ then (7.4.5) is a linear equation. Suppose that $a \neq 0, 1.$ We then define $u(x) = y^{1-a}$ and therefore

$$u^\prime = (1-a) y^\prime y^{-a} = (1-a) (q(x) - p(x) u)$$

or equivalently

$$u^\prime + (1-a) p(x) u = (1-a) q(x),$$ (7.4.6)

a linear equation. For illustration, consider the following example.

EXAMPLE 7.4.6   For $m, n$ constants and $m \neq 0,$ solve $y^\prime - m y + n y^2=0.$
Solution: Let $u = y^{-1}.$ Then $u(x)$ satisfies

$$u^{\prime} + m u = n$$

and its solution is

$$u = A e^{-mx} + e^{-mx} \int n e^{mx} dx = A e^{-mx} + \frac{n}{m}.$$

Equivalently

$$y = \frac{1}{A e^{-mx} + \frac{n}{m}}$$

with $m \neq 0$ and $A$ an arbitrary constant, is the general solution.

EXERCISE 7.4.7  

1. In Example 7.4.6, show that $u^{\prime} + m u = n.$
2. Find the genral solution of the following:
   1. $y^\prime + y = 4.$
   2. $y^\prime - 3 y = 10.$
   3. $y^\prime - 2 x y = 0.$
   4. $y^\prime - x y = 4 x.$
   5. $y^\prime + y = e^{-x}.$
   6. $\sinh x y^\prime + y \cosh x = e^x.$
   7. $(x^2 + 1) y^\prime + 2 x y = x^2.$
3. Solve the following IVP's:
   1. $y^\prime - 4 y = 5, \; y(0) = 0.$
   2. $y^\prime + (1+x^2) y = 3, \; y(0) = 0.$
   3. $y^\prime + y = \cos x, \; y(\pi) = 0.$
   4. $y^\prime- y^2 = 1, \; y(0) = 0.$
   5. $(1+x) y^\prime + y = 2 x^2, \; y(1) = 1.$
4. Let $y_1$ be a solution of $y^\prime + a(x) y = b_1(x)$ and $y_2$ be a solution of $y^\prime + a(x) y = b_2(x).$ Then show that $y_1 + y_2$ is a solution of
   $$y^\prime + a(x) y = b_1(x) + b_2(x).$$

5. Reduce the following to linear equations and hence solve:
   1. $y^\prime + 2 y = y^2.$
   2. $(x y + x^3 e^{y} ) y^\prime = y^2.$
   3. $y^\prime \sin (y) + x \cos (y) = x.$
   4. $y^\prime - y = x y^2.$
6. Find the solution of the IVP
   $$y^\prime + 4 x y + x y^3 = 0, \;\; y(0) = \frac{1}{\sqrt{2}}.$$