# Higher Order ODE IVP

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• Mar 18th 2010, 09:26 PM
mmattson07
Higher 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 PM
Random 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 PM
mmattson07
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}$$\displaystyle (cosx+sinx)+C4e^{-x}(cosx-sinx) \displaystyle y'(0) = 1 so \displaystyle 1=C1+C2-C3+C4 Is this correct so far? • Mar 19th 2010, 01:17 PM Random Variable Are there any other conditions? • Mar 19th 2010, 05:11 PM mmattson07 Yes but I figured since it only gets messier i'd make sure I got this much correct. The other conditions are: \displaystyle y''(0)=0 and \displaystyle y'''(0)=2 I just do what I did above to find the constants? • Mar 19th 2010, 07:29 PM mmattson07 If what I did above is correct then \displaystyle y''(x)= 2C2e^xcosx-2C1e^xsinx+2C3e^{-x}sinx-2C4e^{-x}cosx then \displaystyle 0=2C2-2C4 \displaystyle y'''(x)= 2C2e^x(cosx-sinx)-2C1e^x(sinx+cosx)+2C3e^{-x}$$\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 PM
Random 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 PM
mmattson07
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 PM
Random Variable
Quote:

Originally Posted by mmattson07
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.

I solved the problem using Maple.

Give me a second to solve the problem myself by hand and see what I get.
• Mar 19th 2010, 09:45 PM
Random 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 PM
mmattson07
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 PM
Random Variable
Quote:

Originally Posted by mmattson07
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:

$\displaystyle y''(0)= C1(-2e^0(sin(0))+C2(2e^0(cos(0)))+C3(2e^0(sin(0)))$
$\displaystyle +C4(-2e^0(cos(0)))$
$\displaystyle =>0=0+2C2+0-2C4$
?

Yes. You're correct. My mistake.
• Mar 20th 2010, 12:59 PM
mmattson07
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 PM
Random 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 PM
mmattson07
Ahh, and you can add those because they're both equal to 1?

And the solution to the IVP is $\displaystyle y(x) = (\frac{1}{2}) e^{x} \sin x +(\frac{1}{2})e^{-x} \sin x$
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