Convert 2nd order ode to first order ODES

Aug 2009
21
0
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
 
Apr 2010
384
153
Canada
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 \(\displaystyle y = sin(2t) - \frac{4}{3}cos(2t) \)
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
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...
 
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Apr 2010
384
153
Canada
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!
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
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Apr 2010
384
153
Canada
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.
 
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