Higher Order ODE IVP

**mmattson07** · March 18th 2010, 09:26 PM

Hey MHF,

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

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

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

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

**Random Variable** · March 18th 2010, 09:51 PM

The four roots are $1 \pm i$ and $-1 \pm i.$

So the general solution is $y(x) = C_{1} e^x \cos x + C_{2} e^{x} \sin x + C_{3} e^{-x} \cos x + C_{4}e^{-x} \sin x$

**mmattson07** · March 19th 2010, 12:59 PM

Ok so I can use that to solve the IVP.

$y(0)=0$ so $0 = C1 + C3$
and
$y'(x)= C1e^x(cosx-sinx)+C2e^x(sinx+cosx)-C3e^{-x}$ $(cosx+sinx)+C4e^{-x}(cosx-sinx)$
$y'(0) = 1$ so
$1=C1+C2-C3+C4$

Is this correct so far?

**Random Variable** · March 19th 2010, 01:17 PM

Are there any other conditions?

**mmattson07** · March 19th 2010, 05:11 PM

Yes but I figured since it only gets messier i'd make sure I got this much correct. The other conditions are:

$y''(0)=0$ and
$y'''(0)=2$

I just do what I did above to find the constants?

**mmattson07** · March 19th 2010, 07:29 PM

If what I did above is correct then
$y''(x)= 2C2e^xcosx-2C1e^xsinx+2C3e^{-x}sinx-2C4e^{-x}cosx$ then
$0=2C2-2C4$

$y'''(x)= 2C2e^x(cosx-sinx)-2C1e^x(sinx+cosx)+2C3e^{-x}$ $(cosx-sinx)-2C4e^{-x}(cosx+sinx)$ and
$2= 2C2-2C1+2C3-2C4$

Idk if i'm on the right track here...

**Random Variable** · March 19th 2010, 08:14 PM

There is an error somewhere because the coefficients should be $C_{1}=0, C_{2}= \frac{1}{2}, C_{3} = 0, C_{4} = \frac{1}{2}$ , and I'm not getting that by solving the four equations simultaneously.

**mmattson07** · March 19th 2010, 08:50 PM

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.

**Random Variable** · March 19th 2010, 08:57 PM

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.

**Random Variable** · March 19th 2010, 09:45 PM

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

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

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

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

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

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

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

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

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

which when solved, gives $C_{1}=0, C_{2} = \frac{1}{2}, C_{3} =0, C_{4} = \frac{1}{2}$

**mmattson07** · March 20th 2010, 12:02 PM

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:

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

**Random Variable** · March 20th 2010, 12:09 PM

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:

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

Yes. You're correct. My mistake.

**mmattson07** · March 20th 2010, 12:59 PM

Ok so I had the last sign on y''' wrong yours was correct though. So we have
$y(0)= 0 =C1+C3$
$y'(0)= 1 =C1+C2-C3+C4$
$y''(0)= 0 =2C2-2C4$
$y'''(0)= 2 =-2C1+2C2+2C3+2C4$

How do you deduct the constants from these? The only thing I could see is that
$C2=C4$

**Random Variable** · March 20th 2010, 01:23 PM

From the first equation we also know that $C_{1}=-C_{3}$

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

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

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

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

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

so $C_{1} = 0$ which means that $C_{3}=0$

**mmattson07** · March 20th 2010, 01:46 PM

Ahh, and you can add those because they're both equal to 1?

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

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