-auxiliary eq for y''+9y&=5\cos{5x}

[karush]

\begin{align*}\displaystyle
y''+9y&=5\cos{5x}
\end{align*}
The auxiliary would equation be ??
\begin{align*}\displaystyle
r^2+9&=0\\
%0&\implies r\in\{0,4\}
\end{align*}

 

[MarkFL]

Yes, that's correct...so what would the roots be...and then based on those, what would the homogeneous solution be?

 

[karush]

but are not the roots imaginary?

 

[MarkFL]

but are not the roots imaginary?

Yes...when the characteristic roots are imaginary, such as:

\(\displaystyle r=\pm\beta i\)

then the homogeneous solution takes the form:

\(\displaystyle y_h(x)=c_1\cos(\beta x)+c_2\sin(\beta x)\)

If the roots are complex, such as:

\(\displaystyle r=\alpha\pm\beta i\)

then the homogeneous solution takes the form:

\(\displaystyle y_h(x)=e^{\alpha x}\left(c_1\cos(\beta x)+c_2\sin(\beta x)\right)\)

 

[karush]

very helpful just couldn't find an example like!

 

[MarkFL]

You may be wondering why we get an exponential homogeneous solution when the characteristic roots are real, and sinusoidal when the roots are imaginary...the short story comes from Euler's equation:

\(\displaystyle e^{\beta ix}=\cos(\beta x)+i\sin(\beta x)\)

 

[karush]

I might have to take some pills first
to process that

I assume ...
$$\beta= 3$$

 

[MarkFL]

I might have to take some pills first
to process that

I assume ...
$$\beta= 3$$

 

[karush]

ok so it appears the roots are complex

\begin{align*}\displaystyle
r&=\alpha\pm\beta i\\
&=9\pm3i
\end{align*}
so
\begin{align*}\displaystyle
y_h(x)&=e^{9x}(c_1 \cos(3x) +c_2 \sin(3x))
\end{align*}

really ??

 

[MarkFL]

ok so it appears the roots are complex

\begin{align*}\displaystyle
r&=\alpha\pm\beta i\\
&=9\pm3i
\end{align*}
so
\begin{align*}\displaystyle
y_h(x)&=e^{9x}(c_1 \cos(3x) +c_2 \sin(3x))
\end{align*}

really ??

Let's go back to the characteristic/auxiliary equation:

\(\displaystyle r^2+9=0\)

\(\displaystyle r^2=-9=(3i)^2\)

\(\displaystyle r=\pm3i\)

And so the homogeneous solution is simply:

\(\displaystyle y_h(x)=c_1\cos(3x)+c_2\sin(3x)\)

 

[MarkFL]

The next step is to determine the form the particular solution will take...(Thinking)

 

[karush]

\begin{align*}\displaystyle
&c_1 \cos(3 x)+c_2 \sin(3 x) - \frac{5}{16} \cos(5 x)
\end{align*}

$\textrm{where does the
$\displaystyle- \frac{5}{16} \cos(5 x)$
come from}$

 

[MarkFL]

\begin{align*}\displaystyle
&c_1 \cos(3 x)+c_2 \sin(3 x) - \frac{5}{16} \cos(5 x)
\end{align*}

where does the $- \frac{5}{16} \cos(5 x)$ come from

It comes from either applying the method of undetermined coefficients, the annihilator method, and perhaps other methods I'm not thinking of at the moment. The most straightforward way I know is to use the method of undetermined coefficients.

So, we look at the RHS of the original ODE, and we see it is not a solution to the corresponding homogeneous equation, and so we say the particular solution will take the form:

\(\displaystyle y_p(x)=A\cos(5x)+B\sin(5x)\)

Noting that this is a sinusoid, we know immediately that:

\(\displaystyle y_p''(x)=-25A\cos(5x)-25B\sin(5x)\)

So, we then substitute the particular solution into the original ODE to get:

\(\displaystyle -25A\cos(5x)-25B\sin(5x)+9A\cos(5x)+9B\sin(5x)=5\cos(5x)\)

Arrange this as:

\(\displaystyle -16A\cos(5x)-16B\sin(5x)=5\cos(5x)+0\sin(5x)\)

Equating coefficients, we obtain the system:

\(\displaystyle -16A=5\implies A=-\frac{5}{16}\)

\(\displaystyle -16B=0\implies B=0\)

And so our particular solution is:

\(\displaystyle y_p(x)=-\frac{5}{16}\cos(5x)\)

And then by the principle of superposition, we have the general solution to the ODE:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1\cos(3x)+c_2\sin(3x)-\frac{5}{16}\cos(5x)\)

Now, what if the given ODE had been:

\(\displaystyle y''+9y=5\cos(3x)\)

What would the form for our particular solution be?

 

[karush]

frankly I'm clueless

tried to look at some examples but...

 

[MarkFL]

frankly I'm clueless

tried to look at some examples but...

Well, normally we would say that the particular solution would take the form:

\(\displaystyle y_p=A\cos(3x)+B\sin(3x)\)

But...we should observe that this is our homogeneous solution, and the particular solution needs to be linearly independent. So, what we do is for our particular solution, use the form:

\(\displaystyle y_p=x\left(A\cos(3x)+B\sin(3x)\right)\)

This can be demonstrated to work using the annihilator method, but for now I will leave it at this. You should have a table showing the form to use for the particular solution based on the RHS of a linear inhomogeneous ODE and the homogeneous solution. If not, then I invite you to read this thread:

http://mathhelpboards.com/math-notes-49/justifying-method-undetermined-coefficients-4839.html

 

[karush]

ok, that was very helpful and read the article too.

will open a new OP with new prob
and see if the fog has lifted