Second Order Nonhomogenous Differential Equation

[Legaldose]

Hello everyone, I'm having trouble understanding the solutions to DE's of the form:

ay''+by'+cy=f(t)

We've gone over them in class, I've talked with my friends, and it just doesn't make any sense to me. I was wondering if anyone on here would help me understand the solutions, it would be much appreciated.

 

[vanhees71]

It's a linear equation (supposed that a, \quad b, \quad c are functions of the independent variable x but not of the unknown function y), which makes the task to solve it easier.

First of all, suppose you have two solutions y_1(x) and y_2(x). Now consider y(x)=y_1(x)-y_2(x). Now, because differentiation wrt. x is a linear operation we have
a y''+b y' + c y = (a y_1''+by_1'+c y_1)-(a y_2''+b y_2' +c y_2)=f-f=0.
This means that the difference of two solutions of the inhomogeneous equations is always a solution of the homogeneous equations.

In turn we can conclude that any solution of the inhomogeneous equation is given as the sum of one particular solution of the inhomogeneous equation y_p(x) and the general solution of the homogeneous equation y_h(x).

Further it's clear that with any set of solutions of the homogeneous equation also any linear combination of such solutions is again a solution of the homogeneous equation. The solutions of the homogeneous equation thus build a linear subspace in the vector space of twice differentiable functions. One can prove that this subspace is two-dimensional (look for "Wronskian Determinant" in your textbook or online), i.e., you need to find two linearly independent solutions of the homogeneous equation y_{h1}(x) and y_{h2}(x). Linearly independent means simply that y_{h1}(x)/y_{h2}(x) \neq \text{const}. Then the general solution of the inhomogeneous system reads
y(t)=C_1 y_{h1}(x) + C_2 y_{h2}(x)+y_p(x).
Here, C_1 and C_2 are arbitrary constants.

A solution can be uniquely determined by imposing, e.g., initial conditions,
y(t_0)=y_0, \quad y'(t_0)=y_0'.

 

[Legaldose]

Thank you for spelling this out for me. It already is starting to make more sense. I'll have to spend the next couple of days studying your reply though, as I am busy all weekend. :)