general solution of linear differential equation
is gotten by adding the general solution of the corresponding homogeneous equation
to some particular solution of the nonhomogeneous equation.
The general solution of the homogeneous equation has the form
Example 1. Find the general solution of the nonhomogeneous linear second order differential equation
The corresponding homogeneous equation has apparently the linearly independent solutions and thus the general solution . For finding a particular solution of (2) we variate the constants , , i.e. think that
in the sum
So the second derivative is
Substituting this and the expression of in the differential equation (2) gives the equation
If we then integrate and chose
we can form the particular solution
Accordingly, the general solution of the nonhomogeneous equation (2) is
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
under the boundary conditions
The function is evidently a particular solution of the differential equation. Therefore, the general solution is
Thus we have . By making use of the boundary conditions, we obtain
Solving this system of linear equations and introducing and into the general solution, we have the result
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 equation11Minus sign, on the right-hand member of the equation, it is by convenience in the applications.
The solution of this problem, about boundary values, is known to be given by
where the symmetric function 22Some authors call this symmetry reciprocity’s law. is the so-called Green’s function. It satifies the following boundary problem33It is easy verify the details about such statement; it can be found in any good book on mathematical analysis.
From the last two one, we realize that is continuous at while has there a jump discontinuity. 44The solution , which is above given, it may be physically interpreted as follows: if stands for a displacement and like a force per length unit, then the Green’s function corresponds to a displacement at due a force, of unit magnitude, concentrated at . Let us see an example.
Example 3. Consider the problem
Here, , , , . So from i) and ii), and therefore
Since stays fixed on above Green’s conditions, constants may depend on . Further, the symmetry of demands that , and , where is a constant independent on . Then the continuity condition iii) is automatically satisfied, and the jump condition iv) gives
Thus, the solution is
If, for example, , then we find
In some cases related to partial differential equations (specially that of hyperbolic type), the method of separation of variables, splits in ordinary differential equations (possibly with variable coefficients) on boundary values, and one of them usually leading to a Sturm-Liouville problem (basically an eigen-values and eigen-functions problem). The general solution of those partial differential equations generally leads to Bessel-Fourier series, but the details about that question is out of the sight of this entry.
|Title||general solution of linear differential equation|
|Date of creation||2013-03-22 16:32:25|
|Last modified on||2013-03-22 16:32:25|
|Last modified by||pahio (2872)|
|Synonym||solution of linear ordinary differential equation|