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

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  • Thread starter karush
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In summary,The particular solution to an inhomogeneous ODE will take the form of a sinusoid based on the RHS of the equation and the homogeneous solution.
  • #1
karush
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\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*}
 
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  • #2
Yes, that's correct...so what would the roots be...and then based on those, what would the homogeneous solution be?
 
  • #3
but are not the roots imaginary?
 
  • #4
karush said:
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:

\(\displaystyle y_h(x)=e^{\alpha x}\left(c_1\cos(\beta x)+c_2\sin(\beta x)\right)\)
 
  • #5
very helpful😃 just couldn't find an example like!
 
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  • #6
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:

\(\displaystyle e^{\beta ix}=\cos(\beta x)+i\sin(\beta x)\)
 
  • #7
I might have to take some pills first
to process that😎

I assume ...
$$\beta= 3$$
 
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  • #8
karush said:
I might have to take some pills first
to process that😎

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

65567322.jpg
 
  • #9
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 ??
 
  • #10
karush said:
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:

\(\displaystyle y_h(x)=c_1\cos(3x)+c_2\sin(3x)\)
 
  • #11
The next step is to determine the form the particular solution will take...(Thinking)
 
  • #12
\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}$
 
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  • #13
karush said:
\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?
 
  • #14
frankly I'm clueless

tried to look at some examples but...
 
  • #15
karush said:
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
 
  • #16
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😎
 

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

What is the purpose of an auxiliary equation in a differential equation?

An auxiliary equation is used to find the general solution of a differential equation by converting the original equation into a polynomial equation with constant coefficients.

How is an auxiliary equation derived from a differential equation?

An auxiliary equation is derived by assuming the solution of the differential equation to be in the form of an exponential function, and then substituting it into the original equation and solving for the coefficients.

Why is the auxiliary equation important in solving differential equations?

The auxiliary equation allows us to find the general solution of a differential equation, which is a function that satisfies the original equation for all possible values of the independent variable.

What is the significance of the roots of the auxiliary equation?

The roots of the auxiliary equation represent the values of the independent variable at which the original differential equation has a solution. These roots are also used to determine the type of solution (real, complex, or repeated) for the differential equation.

How does the auxiliary equation help in solving the given differential equation?

By solving the auxiliary equation, we can determine the form of the general solution for the given differential equation. This solution can then be used to find the particular solution by substituting in the initial conditions.

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