The general solution $y$ of the nonhomogeneous linear differential equation $$\frac{d^{n}y}{dx^n}+P_1(x)\frac{d^{n-1}y}{dx^{n-1}}+\cdots+ P_{n-1}(x)\frac{dy}{dx}+P_n(x)y = Q(x)$$ is gotten by adding the general solution $\bar{y}$ of the corresponding homogeneous equation $$\frac{d^{n}\bar{y}}{dx^n}+P_1(x)\frac{d^{n-1}\bar{y}}{dx^{n-1}}+\cdots+ P_{n-1}(x)\frac{d\bar{y}}{dx}+P_n(x)\bar{y} = 0$$ to some particular solution of the nonhomogeneous equation.

The general solution of the homogeneous equation has the form

(1)

Example 1. Find the general solution of the nonhomogeneous linear second order differential equation

(2)

(3)

(4)

In some cases it is not necessary to use the variation of parameters method above illustrated, but a particular solution may be found at simple sight, as it is the case in the following example about boundary values.

Example 2. Find the general solution of the nonhomogeneous linear second order differential equation $$y''-y = 2x$$ under the boundary conditions $$y(1) = 0,\quad y'(0) = 0.$$ The function $x\mapsto -2x$ is evidently a particular solution of the differential equation. Therefore, the general solution is $$y(x) = -2x+C_1e^x+C_2e^{-x}.$$ Thus we have $y'(x) = -2+C_1e^x-C_2e^{-x}$ . By making use of the boundary conditions, we obtain $$0 = y(1) = -2+C_1e+C_2e^{-1},\qquad 0 = y'(0) = -2+C_1-C_2.$$ Solving this system of linear equations and introducing $C_1$ and $C_2$ into the general solution, we have the result $$y(x) = -2x+\frac{2(e+1)}{e^2+1}e^x-\frac{2e(e-1)}{e^2+1}e^{-x}.$$

To solve more advanced problems about nonhomogeneous ordinary linear differential equations of second order with boundary conditions, we may find out a particular solution by using, for instance, the Green's function method. Thus consider, for instance, the self-adjoint differential equation¹ $$\frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)+q(x)y=-f(x), \qquad a<x<b, \qquad y(a)=y(b)=0.$$ The solution of this problem, about boundary values, is known to be given by $$y(x)= \int_a^b G(x,\xi)f(\xi)d(\xi),$$ where the symmetric function $G(x,\xi)=G(\xi,x)$ ² is the so-called Green's function. It satifies the following boundary problem³

Example 3. Consider the problem

Footnotes

...¹

Minus sign, on the right-hand member of the equation, it is by convenience in the applications.

...²

Some authors call this symmetry reciprocity's law.

...³

It is easy verify the details about such statement; it can be found in any good book on mathematical analysis.

...⁴

The solution $y(x)$ , which is above given, it may be physically interpreted as follows: if $y$ stands for a displacement and $f$ like a force per length unit, then the Green's function $G(x,\xi)$ corresponds to a displacement at $x$ due a force, of unit magnitude, concentrated at $x=\xi$ .