#### Trenters4325

##### Junior Member

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- Thread starter Trenters4325
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Let's try \(\displaystyle y''+y'+y=0\)

The characteristic equation is:

\(\displaystyle \L\\m^{2}+m+1=0\)

It's roots are:

\(\displaystyle \L\\\frac{-1}{2}+\frac{\sqrt{3}}{2}i\) and \(\displaystyle \L\\\frac{-1}{2}-\frac{\sqrt{3}}{2}i\)

The -1/2 is \(\displaystyle {\alpha}\)and the\(\displaystyle \pm\frac{\sqrt{3}}{2}\)is \(\displaystyle {\beta}\) in \(\displaystyle {\alpha}+{\beta}i\)

So, you'd have:

\(\displaystyle \L\\y=e^{\frac{-1}{2}x}(C_{1}cos(\frac{\sqrt{3}}{2})x+C_{2}sin(\frac{\sqrt{3}}{2})x)\)

Also,

\(\displaystyle \L\\y=C_{1}e^{(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)x}+C_{2}e^{(\frac{-1}{2}-\frac{\sqrt{3}}{2}i)x}\)

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I can't figure out how the last two equations that you wrote are equivalent. Can you show some of the intermediary steps?

We know are solutions are in the form:

\(\displaystyle \L\\y=C_{1}e^{({\alpha}+i{\beta})x}+C_{2}e^{({\alpha}-i{\beta})x}\)

But, mostly, we would rather have real functions instead of complex

exponentials, so we use Euler's formula:

\(\displaystyle \L\\e^{i{\theta}}=cos{\theta}+isin{\theta}\)

Therefore,

\(\displaystyle \L\\e^{i{\beta}x}=cos{\beta}x+isin{\beta}x\ and\

e^{-i{\beta}x}=cos{\beta}x-isin{\beta}x\)[1]

we have used \(\displaystyle cos({-\beta}x)=cos({\beta}x)\ and\ sin({-\beta}x)={-

sin({\beta}x)\)

By adding and then subtracting the equations in [1], we get:

\(\displaystyle \L\\e^{i{\beta}x}+e^{-i{\beta}x}=2cos{\beta}x\) and

\(\displaystyle \L\\e^{i{\beta}x}-e^{-i{\beta}x}=2isin{\beta}x\)

Since \(\displaystyle \L\\y=C_{1}e^{({\alpha}+i{\beta})x}+C_{2}e^{({\alpha}-i{\beta})x}\) is a solution of \(\displaystyle \L\\ay''+by'+cy=0\)

for any \(\displaystyle C_{1}\ and\ C_{2}\), we can choose \(\displaystyle C_{1}=C_{2}=1\ and\ C_{1}=1,\ C_{2}=-1\) give two solutions:

\(\displaystyle y_{1}=e^{({\alpha}+i{\beta})x}+e^{({\alpha}-i{\beta})x}\) and

\(\displaystyle y_{2}=e^{({\alpha}+i{\beta})x}-e^{({\alpha}-i{\beta})x}\)

But, \(\displaystyle \L\\y_{1}=e^{{\alpha}x}(e^{i{\beta}x}+e^{-i{\beta}x})

=2e^{{\alpha}x}cos{\beta}x\ and\ y_{2}=e^{{\alpha}x}(e^{i{\beta}

x}-e^{-i{\beta}x})=2ie^{{\alpha}x}sin{\beta}x\)

The superposition principle says that the real functions \(\displaystyle \L\\e^

{{\alpha}x}cos{\beta}x\ and\ e^{{\alpha}x}sin{\beta}x\) are

solutions of

\(\displaystyle \L\\ay''+by'+cy=0\)

So, the general solution is:

\(\displaystyle \L\\y=C_{1}e^{{\alpha}x}cos{\beta}x+C_{2}e^{{\alpha}x}sin

{\beta}x=e^{{\alpha}x}(C_{1}cos{\beta}x+C_{2}sin{\beta}x)\)

I hope this helps. If you need a more thorough explanation, I'd suggest

seeing your professor. I'm LaTexed out :lol:

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How did you get rid of the imaginary component in the coefficient of that sine function?galactus said:\(\displaystyle =2e^{{\alpha}x}cos{\beta}x\ and\ y_{2}=e^{{\alpha}x}(e^{i{\beta}

x}-e^{-i{\beta}x})=2ie^{{\alpha}x}sin{\beta}x\)

The superposition principle says that the real functions \(\displaystyle \L\\e^

{{\alpha}x}cos{\beta}x\ and\ e^{{\alpha}x}sin{\beta}x\) are

solutions of

\(\displaystyle \L\\ay''+by'+cy=0\)

So, the general solution is:

\(\displaystyle \L\\y=C_{1}e^{{\alpha}x}cos{\beta}x+C_{2}e^{{\alpha}x}sin\)

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So, you multiplied it by 1/2i?