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#### karush

##### Well-known member

- Jan 31, 2012

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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*}

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

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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

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- #3

- Jan 31, 2012

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but are not the roots imaginary?

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- #4

Yes...when the characteristic roots are imaginary, such as:but are not the roots imaginary?

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

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:

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

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- #5

- Jan 31, 2012

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very helpful just couldn't find an example like!

Last edited:

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- #6

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

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- #7

- Jan 31, 2012

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I might have to take some pills first

to process that

I assume ...

$$\beta= 3$$

to process that

I assume ...

$$\beta= 3$$

Last edited:

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- #8

I might have to take some pills first

to process that

I assume ...

$$\beta= 3$$

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- #9

- Jan 31, 2012

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\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 ??

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- #10

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

\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 ??

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

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

\(\displaystyle r=\pm3i\)

And so the homogeneous solution is simply:

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

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- #11

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- #12

- Jan 31, 2012

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\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}$

&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:

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- #13

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.\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

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?

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- #14

- Jan 31, 2012

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frankly I'm clueless

tried to look at some examples but...

tried to look at some examples but...

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- #15

Well, normally we would say that the particular solution would take the form:frankly I'm clueless

tried to look at some examples but...

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

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

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- #16

- Jan 31, 2012

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will open a new OP with new prob

and see if the fog has lifted