Could you please help me with this kind of question?Any respond would be greatly thanks

http://i270.photobucket.com/albums/j...untitled-6.jpg

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- Jun 14th 2008, 10:15 PMnoobSolve Differential Equation..Urgent Help Needed
Could you please help me with this kind of question?Any respond would be greatly thanks

http://i270.photobucket.com/albums/j...untitled-6.jpg - Jun 14th 2008, 10:17 PMnoob
- Jun 14th 2008, 10:50 PMChris L T521
- Jun 14th 2008, 11:01 PMnoob
- Jun 14th 2008, 11:04 PMnoob
- Jun 14th 2008, 11:14 PMnoob
- Jun 14th 2008, 11:14 PMChris L T521
Doesn't this look like the Characteristic equation? (It is...)

Since our C.E. is $\displaystyle r^4+2r^3+3r^2+2r+1 $, it factored out to $\displaystyle \bigg(x^2+x+1\bigg)^2$.

Thus $\displaystyle r=\frac{-1\pm\sqrt{1-4}}{2}=\bigg(-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\bigg)$ with multiplicity 2...

Can you find the general solution from here?

--Chris - Jun 14th 2008, 11:25 PMnoob
- Jun 14th 2008, 11:45 PMChris L T521
Since the conjugate roots are repeated, the general solution takes the form:

$\displaystyle y=c_1e^{-\frac{1}{2}+\frac{\sqrt{3}}{2}i}+c_2e^{-\frac{1}{2}-\frac{\sqrt{3}}{2}i}+c_3xe^{-\frac{1}{2}+\frac{\sqrt{3}}{2}i}+c_4xe^{-\frac{1}{2}-\frac{\sqrt{3}}{2}i}$

However, we don't like to see complex numbers in the solution, so we use**Euler's Formula**$\displaystyle \bigg(e^{i\theta}=\cos(\theta)+i\sin(\theta)\bigg)$ to clean things up:

After some calculations, we find that our general solution is:

$\displaystyle y=e^{-\frac{1}{2}x}\left[c_1\cos\left(\frac{\sqrt{3}}{2}\right)+c_2\sin\lef t(\frac{\sqrt{3}}{2}\right)\right]+xe^{-\frac{1}{2}x}\left[c_3\cos\left(\frac{\sqrt{3}}{2}\right)+c_4\sin\lef t(\frac{\sqrt{3}}{2}\right)\right]$

$\displaystyle \implies \color{red}\boxed{y=e^{-\frac{1}{2}x}\left[\left(c_1+c_3x\right)\cos\left(\frac{\sqrt{3}}{2}\ right)+\left(c_2+c_4x\right)\sin\left(\frac{\sqrt{ 3}}{2}\right)\right]}$

To see how we go from $\displaystyle y=c_1e^{(\alpha+\beta i)x}+c_2e^{(\alpha-\beta i) x}\implies y=e^{\alpha x}\left[c_1\cos(\beta x)+c_2\sin(\beta x)\right]$, read post #3 in my Diff Equ Tutorial. - Jun 15th 2008, 12:01 AMnoob