# Thread: Convert 2nd order ode to first order ODES

1. ## Convert 2nd order ode to first order ODES

Hello,
I have the DE:

$\displaystyle y''(t) + y = 0$

with IC's:

$\displaystyle y'(0) = 1$hs

$\displaystyle y(0) = 0$

If I convert this to a system of 1st order ODES is it just:

$\displaystyle \frac{dy_1}{dt} = y_2, \ \ \ \frac{dy_2}{dt} = -y_1$

How about if the DE is:

$\displaystyle y''(t) + y = \sin(2t)$ and the same IC's.

Thanks

2. Originally Posted by RoyalFlush
Hello,
I have the DE:

$\displaystyle y''(t) + y = 0$

with IC's:

$\displaystyle y'(0) = 1$hs

$\displaystyle y(0) = 0$

If I convert this to a system of 1st order ODES is it just:

$\displaystyle \frac{dy_1}{dt} = y_2, \ \ \ \frac{dy_2}{dt} = -y_1$

How about if the DE is:

$\displaystyle y''(t) + y = \sin(2t)$ and the same IC's.

Thanks
The first question is a second order homogenious equation. We find the roots by computing the characteristic equation.

$\displaystyle \lambda^2 + 1 = 0$

Clearly this has the imaginary roots $\displaystyle -i$ and $\displaystyle i$

So,

$\displaystyle y = c_1 e^{-i x} + c_2 e^{ i x }$

Compute and sub in initial conditions.

The second question can be computed by method of comparing co-efficients (we can also use variation of paramters but co-efficients is easiast here).

For $\displaystyle y''(t) + y = \sin(2t)$ the general solution is the solution from question 1 plus the particular solution.

For the particular solution,

Let $\displaystyle y = Asin(2t) + Bcos(2t)$

So,

$\displaystyle y^{ \prime } = 2Acos(2t) -2Bsin(2t)$

$\displaystyle y^{ \prime \prime } = -4Acos(2t) - 4Bcos(2t)$

Subbing into the equation we get,

$\displaystyle -4Acos(2t) - 4Bcos(2t) + Asin(2t) + Bcos(2t) = sin(2t)$

$\displaystyle cos(2t) [ - 4A - 3B ] + Asin(2t) = sin(2t)$

$\displaystyle A = 1$

$\displaystyle -4 -3B = 0$

$\displaystyle B = - \frac{4}{3}$

And $\displaystyle y = sin(2t) - \frac{4}{3}cos(2t)$

3. Originally Posted by AllanCuz
The first question is a second order homogenious equation. We find the roots by computing the characteristic equation.

$\displaystyle \lambda^2 + 1 = 0$

Clearly this has the imaginary roots $\displaystyle -i$ and $\displaystyle i$

So,

$\displaystyle y = c_1 e^{-i x} + c_2 e^{ i x }$

Compute and sub in initial conditions.
You should really use Euler's Formula to clean this up...

4. Originally Posted by Prove It
You should really use Euler's Formula to clean this up...
You mean, actually do work? Pshh...

$\displaystyle y = c_1 e^{ -ix} + c_2 e^{ix}$

$\displaystyle = c_1 [ cosx - isinx ] + c_2 [cosx + isinx]$

$\displaystyle = [c_1 + c_2]cosx + [ i c_2 - i c_1 ] sinx$

Let $\displaystyle A = c_1 + c_2$ and $\displaystyle B = i c_2 - i c_2$

$\displaystyle y = Acosx + Bsinx$

This is actually the result of a seperation of variables of a wave equation :O!

5. That's very clever, guys, but it doesn't actually answer the question asked, does it?

Royal Flush, yes, if you let $\displaystyle y_1= y$ and $\displaystyle y_2= \frac{dy}{dt}$, then we have immediately that $\displaystyle \frac{dy_1}{dt}= y_2$ and then $\displaystyle \frac{d^2y}{dt^2}= \frac{d}{dt}\frac{dy_1}{dt}= \frac{dy_2}{dt}$ so that $\displaystyle \frac{d^2y}{dt^2}+ y= \frac{dy_2}{dt}+ y_1= 0$ so that $\displaystyle \frac{dy_2}{dt}= -y_1$

And your initial conditions are $\displaystyle y_1(0)= 0$ and $\displaystyle y_2(0)=1$.

This can also be written as the "matrix" equation:
$\displaystyle \frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dt}= \begin{pmatrix}y_2 \\ -y_1\end{pmatrix}= \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}$ subject to the initial condition $\displaystyle \begin{pmatrix}y_1(0) \\ y_2(0)\end{pmatrix}= \begin{pmatrix}0 \\ 1\end{pmatrix}$.

In the same way, the equation y"+ y= sin(2t) becomes the pair of equations $\displaystyle \frac{dy_1}{dt}= y_2$ and $\displaystyle \frac{dy_2}{dt}= -y_1+ sin(2t)$ with the same intial conditions.

That also can be written as a "matrix" equation:
$\displaystyle \frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dt}=$$\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}+ \begin{pmatrix}0 \\ sin(2t)\end{pmatrix} with initial condition \displaystyle \begin{pmatrix}y_1(0) \\ y_2(0)\end{pmatrix}= \begin{pmatrix}0 \\ 1\end{pmatrix}. Of course, the eigenvalues of the matrix \displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} are i and -i which lead to the complex exponentials (sine and cosine) solutions Prove It and AllenCuz show. 6. Originally Posted by HallsofIvy That's very clever, guys, but it doesn't actually answer the question asked, does it? Royal Flush, yes, if you let \displaystyle y_1= y and \displaystyle y_2= \frac{dy}{dt}, then we have immediately that \displaystyle \frac{dy_1}{dt}= y_2 and then \displaystyle \frac{d^2y}{dt^2}= \frac{d}{dt}\frac{dy_1}{dt}= \frac{dy_2}{dt} so that \displaystyle \frac{d^2y}{dt^2}+ y= \frac{dy_2}{dt}+ y_1= 0 so that \displaystyle \frac{dy_2}{dt}= -y_1 And your initial conditions are \displaystyle y_1(0)= 0 and \displaystyle y_2(0)=1. This can also be written as the "matrix" equation: \displaystyle \frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dt}= \begin{pmatrix}y_2 \\ -y_1\end{pmatrix}= \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix} subject to the initial condition \displaystyle \begin{pmatrix}y_1(0) \\ y_2(0)\end{pmatrix}= \begin{pmatrix}0 \\ 1\end{pmatrix}. In the same way, the equation y"+ y= sin(2t) becomes the pair of equations \displaystyle \frac{dy_1}{dt}= y_2 and \displaystyle \frac{dy_2}{dt}= -y_1+ sin(2t) with the same intial conditions. That also can be written as a "matrix" equation: \displaystyle \frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dt}=$$\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}+ \begin{pmatrix}0 \\ sin(2t)\end{pmatrix}$
with initial condition $\displaystyle \begin{pmatrix}y_1(0) \\ y_2(0)\end{pmatrix}= \begin{pmatrix}0 \\ 1\end{pmatrix}$.

Of course, the eigenvalues of the matrix $\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ are i and -i which lead to the complex exponentials (sine and cosine) solutions Prove It and AllenCuz show.
He said "if I convert to" which I took to mean he was trying to solve the problem via that method, and it wasn't specifically asked of him.