Hey MHF,

I got stuck on finding the gen sol. to the following ODE

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

I solved the auxiliary eqn. $\displaystyle m^4+4= 0$ for $\displaystyle m= ^4\sqrt{-4}$

So the roots must be complex? This is where I got stuck.

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- Mar 18th 2010, 09:26 PMmmattson07Higher Order ODE IVP
Hey MHF,

I got stuck on finding the gen sol. to the following ODE

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

I solved the auxiliary eqn. $\displaystyle m^4+4= 0$ for $\displaystyle m= ^4\sqrt{-4}$

So the roots must be complex? This is where I got stuck. - Mar 18th 2010, 09:51 PMRandom Variable
The four roots are $\displaystyle 1 \pm i $ and $\displaystyle -1 \pm i. $

So the general solution is $\displaystyle y(x) = C_{1} e^x \cos x + C_{2} e^{x} \sin x + C_{3} e^{-x} \cos x + C_{4}e^{-x} \sin x $ - Mar 19th 2010, 12:59 PMmmattson07
Ok so I can use that to solve the IVP.

$\displaystyle y(0)=0$ so $\displaystyle 0 = C1 + C3$

and

$\displaystyle y'(x)= C1e^x(cosx-sinx)+C2e^x(sinx+cosx)-C3e^{-x}$$\displaystyle (cosx+sinx)+C4e^{-x}(cosx-sinx)$

$\displaystyle y'(0) = 1$ so

$\displaystyle 1=C1+C2-C3+C4$

Is this correct so far? - Mar 19th 2010, 01:17 PMRandom Variable
Are there any other conditions?

- Mar 19th 2010, 05:11 PMmmattson07
Yes but I figured since it only gets messier i'd make sure I got this much correct. The other conditions are:

$\displaystyle y''(0)=0$ and

$\displaystyle y'''(0)=2$

I just do what I did above to find the constants? - Mar 19th 2010, 07:29 PMmmattson07
If what I did above is correct then

$\displaystyle y''(x)= 2C2e^xcosx-2C1e^xsinx+2C3e^{-x}sinx-2C4e^{-x}cosx$ then

$\displaystyle 0=2C2-2C4$

$\displaystyle y'''(x)= 2C2e^x(cosx-sinx)-2C1e^x(sinx+cosx)+2C3e^{-x}$$\displaystyle (cosx-sinx)-2C4e^{-x}(cosx+sinx)$ and

$\displaystyle 2= 2C2-2C1+2C3-2C4$

Idk if i'm on the right track here... - Mar 19th 2010, 08:14 PMRandom Variable
There is an error somewhere because the coefficients should be $\displaystyle C_{1}=0, C_{2}= \frac{1}{2}, C_{3} = 0, C_{4} = \frac{1}{2} $, and I'm not getting that by solving the four equations simultaneously.

- Mar 19th 2010, 08:50 PMmmattson07
I know y' is correct, I will double check y'' and y'''. Those roots make all four equations true though, how did you find them without solving the four equations simultaneously? Thanks.

- Mar 19th 2010, 08:57 PMRandom Variable
- Mar 19th 2010, 09:45 PMRandom Variable
$\displaystyle y(x) = C_{1} e^x \cos x + C_{2} e^{x} \sin x + C_{3} e^{-x} \cos x + C_{4}e^{-x} \sin x $

$\displaystyle y'(x) = C_{1} (e^{x} \cos x - e^{x} \sin x) + C_{2} (e^{x} \sin x + e^{x} \cos x) + C_{3} (-e^{-x} \cos x - e^{-x} \sin x) + C_{4} $ $\displaystyle (-e^{-x} \sin x+ e^{-x} \cos x) $

$\displaystyle y''(x) = C_{1}(e^{x} \cos x -e^{x} \sin x -e^{x} \sin x - e^{x} \ \cos x) + C_{2}( e^{x}\sin x + e^{x}\cos x + e^{x} \cos x - e^{x} $ $\displaystyle \sin x) + C_{3} (e^{-x}\cos x + e^{-x} \sin x+ e^{-x} \sin x - e^{-x} \cos x) + C_{4} $ $\displaystyle ( e^{-x} \sin x - e^{-x} \cos x -e^{-x} \cos x - e^{-x} \sin x)$

$\displaystyle = -2C_{1}(e^{x}\sin x) + 2 C_{2}e^{x}\cos x + 2C_{3}e^{-x}\sin x -2C_{4} e^{-x} \cos x $

$\displaystyle y'''(x) = -2C_{1}(e^{x} \sin x + e^{x} \cos x) + 2C_{2}(e^{x} \cos x - e^{x} \sin x) + 2C_{3} (-e^{-x} \sin x +e^{-x} \cos x) $ $\displaystyle - 2C_{4} (-e^{-x}\cos x - e^{-x} \sin x) $

$\displaystyle y(0) = 0 = C_{1}+ C_{3} $

$\displaystyle y'(0) = 1= C_{1} + C_{2} - C_{3} + C_{4} $

EDIT: $\displaystyle y''(0) = 0 = 2C_{2} - 2C_{4} $

$\displaystyle y'''(0) = 2 = -2 C_{1} + 2C_{2} +2C_{3} + 2C_{4} $

which when solved, gives $\displaystyle C_{1}=0, C_{2} = \frac{1}{2}, C_{3} =0, C_{4} = \frac{1}{2} $ - Mar 20th 2010, 12:02 PMmmattson07
Your y'' matches the one I have above. However isn't sin(0)=0? So the C1 and C3 terms would go away when evaluated at 0? Leaving:

$\displaystyle y''(0)= C1(-2e^0(sin(0))+C2(2e^0(cos(0)))+C3(2e^0(sin(0)))$

$\displaystyle +C4(-2e^0(cos(0)))$

$\displaystyle =>0=0+2C2+0-2C4$

? - Mar 20th 2010, 12:09 PMRandom Variable
- Mar 20th 2010, 12:59 PMmmattson07
Ok so I had the last sign on y''' wrong yours was correct though. So we have

$\displaystyle y(0)= 0 =C1+C3$

$\displaystyle y'(0)= 1 =C1+C2-C3+C4$

$\displaystyle y''(0)= 0 =2C2-2C4$

$\displaystyle y'''(0)= 2 =-2C1+2C2+2C3+2C4$

How do you deduct the constants from these? The only thing I could see is that

$\displaystyle C2=C4$ - Mar 20th 2010, 01:23 PMRandom Variable
From the first equation we also know that $\displaystyle C_{1}=-C_{3}$

Then from the second equation $\displaystyle 2C_{1}+2C_{2} =1 $ (1)

And from the fourth equation $\displaystyle -4C_{1}+4C_{2}=2 $ or $\displaystyle -2C_{1}+2C_{2}=1 $ (2)

Add (1) and (2) to get $\displaystyle 4C_{2} = 2 $

so $\displaystyle C_{2} = \frac{1}{2}$ which means that $\displaystyle C_{4}=\frac{1}{2} $

and going back to (1), $\displaystyle 2C_{1}+2\Big(\frac{1}{2}\Big) = 1 $

so $\displaystyle C_{1} = 0$ which means that $\displaystyle C_{3}=0 $ - Mar 20th 2010, 01:46 PMmmattson07
Ahh, and you can add those because they're both equal to 1?

And the solution to the IVP is $\displaystyle

y(x) = (\frac{1}{2}) e^{x} \sin x +(\frac{1}{2})e^{-x} \sin x

$