# Linear system of equations for a differential equation

1. May 11, 2013

### Abyssnight

Solve the equation $m\frac{d^{2}x}{dt^{2}} + c\frac{dx}{dt} + kx = (ax + b)^{2} + c^{2}$ for the constants $m, c, k$

The right hand side a, b, and c are arbitrary digits. For me they are a = 2, b = 3, and c = 8.
The problem recommends creating a linear system of equations for me to solve. This is to be done using MapleSoft.

I tried making a set of equations using (m*r^2 + c*r + k) = (2+3)^2 + 64 and then using various values of r to get a system of equations. But the result got me $m = 0, c = 0$ and k equaling some number. However this is incorrect because the left part of the main equations with the constants will be used to make a vibration model that I would solve. With m and c equaling 0, I would not have any derivatives and therefore not being able to solve the equation.

So now I'm at a loss. I also tried converting the equation into 2 systems of first order differentials but that lead to me really nowhere.

2. May 11, 2013

### HallsofIvy

Staff Emeritus
I'm afraid you are completely confused as to what you are doing. You titled this "linear system of equations" but this cannot be put into "linear system" form because your equation is NOT linear to begin with: x(t) is the dependent variable and you have a non-linear function of the form $(ax+ b)^2$.

A linear equation would have to have a function of the independent variable, t, on the right, not a function of the dependent variable x.

That is $m d^2x/dt^2+ c dx/dt+ kx= (at+ )^2+ c$ would be a linear equation, not the equation you have.

The standard method of solution would be to solve the corresponding homogeneous equation first: solve the characteristic equation $mr^2+ cr+ k= 0$ to get a general solution to the associated homogeneous equation then try to find a solution of the form "$x= At^2+ Bt+ C$".

I'm not sure what you mean by "using various values of r to get a system of equations." You need to get a system of first order differential equations from the orginal equation. To do that defined a new dependent variable, y(t), as the derivative of x: y(t)= dx/dt. Then the second derivative of x is just the first derivative of y and "$md^2x/dt+ cdx/dt+ kx$" becomes "$m dy/dt+ c dx/dt+ kx$" and the entire differential equation becomes
$mdy/dt+ c dx/dt+ kx= (at+ b)^2+ c$
and, of course, $dx/dt- y= 0$

That could be written as a matrix equation:
$$m\frac{\begin{pmatrix}x \\ y\end{pmatrix}}{dt}+ \begin{pmatrix}k & c \\ 0 & -1 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}(at+ b)^2+ c \\ 0 \end{pmatrix}$$

Last edited: May 11, 2013