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[SOLVED] The auxiliary would equation be

karush

Well-known member
Jan 31, 2012
2,928
\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

Administrator
Staff member
Feb 24, 2012
13,775
Yes, that's correct...so what would the roots be...and then based on those, what would the homogeneous solution be?
 

karush

Well-known member
Jan 31, 2012
2,928
but are not the roots imaginary?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,928
very helpful😃 just couldn't find an example like!
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,928
I might have to take some pills first
to process that😎

I assume ...
$$\beta= 3$$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

karush

Well-known member
Jan 31, 2012
2,928
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Administrator
Staff member
Feb 24, 2012
13,775
The next step is to determine the form the particular solution will take...(Thinking)
 

karush

Well-known member
Jan 31, 2012
2,928
\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}$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\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

Well-known member
Jan 31, 2012
2,928
frankly I'm clueless

tried to look at some examples but...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

Justifying the Method of Undetermined Coefficients
 

karush

Well-known member
Jan 31, 2012
2,928
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😎