Hello,

Given is $\displaystyle y^{4}+4y=0$ Who can I solove this? Who can do this step by step?

Greets Thanks .

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- Jun 25th 2006, 10:35 AMBertDe.
Hello,

Given is $\displaystyle y^{4}+4y=0$ Who can I solove this? Who can do this step by step?

Greets Thanks . - Jun 25th 2006, 01:09 PMJameson
By the title "DE", is this a differential equation? Does your equation mean y''''+4y=0?

- Jun 25th 2006, 04:38 PMThePerfectHackerQuote:

Originally Posted by**Bert**

$\displaystyle y^{(4)}+4y=0$

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If,

$\displaystyle y^{(4)}+4y=0$

The charachteristic equation is,

$\displaystyle k^4+4=0$

Thus,

$\displaystyle k^4=-4$

The solutions are,

$\displaystyle k=1+i,1-i,-1+i,-1-i$

Since the solutions are complex conjakates thus the general solution is,

$\displaystyle y=C_1e^x(\cos x+\sin x)+C_2e^x(\cos x-\sin x)$$\displaystyle +C_3e^{-x}(\cos x+\sin x)+C_4e^{-x}(\cos x-\sin x)$ - Jun 26th 2006, 12:15 AMBert
who do you get those complex solotions?

- Jun 26th 2006, 02:18 AMBert
can you put $\displaystyle \frac{\sqrt{2}}{2}$ into the ct?

- Jun 26th 2006, 10:36 AMThePerfectHackerQuote:

Originally Posted by**Bert**

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Method 1) (Informal but it works!)

$\displaystyle k^4=-4$

Thus, (square root),

$\displaystyle k^2=\pm 2 i$

If $\displaystyle k^2=2i$

Take square roots,

$\displaystyle k=\pm \sqrt{2}\sqrt{i}$

But, $\displaystyle \sqrt{i}=\sqrt{2}/2\pm \sqrt{2}/2 i$Thus,

$\displaystyle k=\pm \sqrt{2} (\sqrt{2}/2\pm\sqrt{2}/2 i)$

There are hence 4 possibilities (based on plus minus signs).

$\displaystyle k=-1-i,1+i,1-i,-1+i$

Finally if,

$\displaystyle k^2=-2i$ then by the same argument you get the same solutions, as before. - Jun 26th 2006, 10:48 AMThePerfectHacker
Method 2) (The is the elegant way. [My method if you are curious was to guess and play around with the roots cuz I was too lazy to work it out])

You have,

$\displaystyle k^4=-4$

Express this in polar form,

$\displaystyle k^4=4(\cos \pi+i\sin \pi)$

Now apply, de Moivre's theorem

And the roots are,

$\displaystyle 4^{1/4} \left( \cos \left( \frac{\pi}{4}+\frac{2\pi k}{4}\right) +i\sin \left( \frac{\pi}{4}+\frac{2\pi k}{4} \right) \right)$ one for each value of $\displaystyle k=0,1,2,3$. And $\displaystyle 4^{1/4}=(4^{1/2})^{1/2}=2^{1/2}=\sqrt{2}$

You get your solutions.

$\displaystyle \sqrt{2} (\cos \pi/4+i\sin \pi/4)=1+i$

$\displaystyle \sqrt{2} (\cos 3\pi/4+i\sin 3\pi/4)=-1+i$

$\displaystyle \sqrt{2} (\cos 5\pi/4+i\sin 5\pi/4)=-1-i$

$\displaystyle \sqrt{2} (\cos 7\pi/4+i\sin 7\pi/4)=-1+i$ - Jun 26th 2006, 12:54 PMBert
O yes thenk you very wel I see Greets.