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

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Discussion Overview

The discussion revolves around solving the differential equation \(y'' + 9y = 5\cos(5x)\), focusing on the auxiliary equation, the nature of its roots, and the forms of the homogeneous and particular solutions. Participants explore the implications of imaginary roots and the methods for determining particular solutions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant proposes the auxiliary equation \(r^2 + 9 = 0\) and questions the nature of its roots.
  • Another participant confirms that the roots are indeed imaginary, leading to a discussion about the form of the homogeneous solution.
  • Several participants discuss the implications of having complex roots, with one stating that the homogeneous solution takes the form \(y_h(x) = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))\) when roots are complex.
  • A participant suggests that the particular solution should take the form \(y_p(x) = A\cos(5x) + B\sin(5x)\) based on the method of undetermined coefficients.
  • Another participant elaborates on the process of finding the coefficients for the particular solution, leading to the conclusion that \(A = -\frac{5}{16}\) and \(B = 0\).
  • There is a discussion about what the form of the particular solution would be if the right-hand side of the equation were \(5\cos(3x)\), with one participant expressing confusion about the approach.
  • Another participant explains that the particular solution must be linearly independent from the homogeneous solution, suggesting a modified form involving \(x\).

Areas of Agreement / Disagreement

Participants generally agree on the nature of the roots being complex and the forms of the homogeneous solution. However, there is some uncertainty regarding the approach to finding the particular solution, especially when the right-hand side matches the homogeneous solution.

Contextual Notes

There are unresolved aspects regarding the application of methods for determining particular solutions, particularly in cases where the right-hand side is similar to the homogeneous solution. Some participants reference methods like the annihilator method without fully detailing them.

Who May Find This Useful

This discussion may be useful for students and practitioners dealing with ordinary differential equations, particularly those interested in methods for solving linear inhomogeneous equations and understanding the implications of complex roots.

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*}
 
Last edited:
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Yes, that's correct...so what would the roots be...and then based on those, what would the homogeneous solution be?
 
but are not the roots imaginary?
 
karush said:
but are not the roots imaginary?

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

$$r=\pm\beta i$$

then the homogeneous solution takes the form:

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

If the roots are complex, such as:

$$r=\alpha\pm\beta i$$

then the homogeneous solution takes the form:

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

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

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

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

65567322.jpg
 
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:

$$r^2+9=0$$

$$r^2=-9=(3i)^2$$

$$r=\pm3i$$

And so the homogeneous solution is simply:

$$y_h(x)=c_1\cos(3x)+c_2\sin(3x)$$
 
  • #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}$
 
Last edited:
  • #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:

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

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

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

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

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

Arrange this as:

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

Equating coefficients, we obtain the system:

$$-16A=5\implies A=-\frac{5}{16}$$

$$-16B=0\implies B=0$$

And so our particular solution is:

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

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

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

Now, what if the given ODE had been:

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

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:

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

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:

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

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😎
 

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