# To solve ar''+br'+cr=0

1. Jun 11, 2014

### Jhenrique

Exist solution for ar''(t) + br'(t) + cr(t) = 0 ? If yes how is?

PS: r is a vector.

2. Jun 11, 2014

### Matterwave

That form of differential equation is basically for a damped harmonic oscillator: http://en.wikipedia.org/wiki/Damped_harmonic_oscillator#Damped_harmonic_oscillator.

The fact that [bf]r[/bf] is a vector is not really important to this problem since the problem decouples to 3 identical equations for x y and z (there's no term in that equation that couples the different directions to each other).

Solving just for x (the solutions for y and z are exactly the same): Try the solution $x=e^{\lambda t}$, then this equation becomes:

$$a\lambda^2 e^{\lambda t}+b\lambda e^{\lambda t}+c e^{\lambda t}=0$$

Canceling the common factor $e^{\lambda t}$ we get:

$$a\lambda^2+b\lambda+c=0$$

This is a quadratic equation for $\lambda$ with the usual solution:

$$\lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The differential equation is linear so we can add two different solutions together and still get a solution. The general solution is then:

$$x=A \exp\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)+B\exp\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$$

More detail can be found here: http://en.wikipedia.org/wiki/Damping

3. Jun 11, 2014

### Jhenrique

The fact of the differential equation be vectorial and not scalar changes everything!

The solution for $y'(x)=ky(x)$ is $y=C \exp(kx)$ and graphically is an exponential curve, but the solution for $\vec{r}'(t) = K \vec{r}(t)$ is $\vec{r} = C_1 \exp(\lambda_1 t) \hat{v}_1 + C_2 \exp(\lambda_2 t) \hat{v}_2$ and graphically is or a saddle, or a node, or a proper node, etc...

Realize that my doubt, in matrix terms, will be so:
$$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \begin{bmatrix} x''\\ y''\\ \end{bmatrix} + \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix} \begin{bmatrix} x'\\ y'\\ \end{bmatrix} + \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ \end{bmatrix} \begin{bmatrix} x\\ y\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$

4. Jun 11, 2014

### HallsofIvy

Well, you do what Matterwave suggested just taking those coefficients to be matrices instead of numbers. All of those operations, exponential, square root, etc. can be defined for matrices.

5. Jun 11, 2014

### Staff: Mentor

Another way a problem like this is sometimes done is to define a new vector x by:

x2n-1=rn
x2n=r'n

such that dx2n-1/dt=x2n

Then you can rewrite the differential equation in the form dx/dt = Ax, where A is a 2n x 2n matrix. Then, when you make the substitution x = ceλt, it leads to an ordinary eigenvalue value problem (A-λI)x = 0.

Chet