# Second Order Nonhomogenous Differential Equation

1. Nov 1, 2013

### 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.

2. Nov 1, 2013

### 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,
$$y(t_0)=y_0, \quad y'(t_0)=y_0'.$$

3. Nov 2, 2013

### 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. :)