# Math Help - Solve Differential Equation..Urgent Help Needed

1. ## Solve Differential Equation..Urgent Help Needed

2. Originally Posted by noob

How can i factor the equation??

3. Originally Posted by noob
How can i factor the equation??
Hint : Expand $\bigg(r^2+r+1\bigg)^2$...

--Chris

4. Originally Posted by Chris L T521
Hint : Expand $\bigg(r^2+r+1\bigg)^2$...

--Chris
normally as to factorized the equation, i'll use synthetic division, but in that case, i'll not be able to find a single value that satisfying the equation (try and error method)

5. Originally Posted by Chris L T521
Hint : Expand $\bigg(r^2+r+1\bigg)^2$...

--Chris
is it we get : $r^4+2r^3+3r^2+2r+1$ ????

6. Originally Posted by Chris L T521
Hint : Expand $\bigg(r^2+r+1\bigg)^2$...

--Chris

i've got an imaginary number as their factor

$
(r^2+r+1)^2$

[(-0.5+0.866i)(-0.5-0.866i)]^2

how can i solved it??i really confused rite now

7. Originally Posted by noob
is it we get : $r^4+2r^3+3r^2+2r+1$ ????
Doesn't this look like the Characteristic equation? (It is...)

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

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

Can you find the general solution from here?

--Chris

8. Originally Posted by Chris L T521
Doesn't this look like the Characteristic equation? (It is...)

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

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

Can you find the general solution from here?

--Chris
is is the general solution lokk like this

y(x) = c1 e^(1/2 + surd3/2 i ) + c2 e^(1/2 - surd3/2 i)

the other two factor is what??i might done it wrongly..correct me plese..thanks in advanced

9. Originally Posted by noob
is is the general solution lokk like this

y(x) = c1 e^(1/2 + surd3/2 i ) + c2 e^(1/2 - surd3/2 i)

the other two factor is what??i might done it wrongly..correct me plese..thanks in advanced
Since the conjugate roots are repeated, the general solution takes the form:

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

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

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

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

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

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

10. Originally Posted by Chris L T521
Since the conjugate roots are repeated, the general solution takes the form:

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

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

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

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

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

To see how we go from $y=c_1e^{(\alpha+\beta i)x}+c_2e^{(\alpha-\beta i) x}\implies y=e^{\alpha x}\left[c_1\cos(\beta x)+c_2\sin(\beta x)\right]$, read post #3 in my Diff Equ Tutorial.
thanks mates..thank you very much..still looking and try to understand the calculation