# Thread: Higher Order ODE IVP

1. ## 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.

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

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

3. 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}$$\displaystyle (cosx+sinx)+C4e^{-x}(cosx-sinx) \displaystyle y'(0) = 1 so \displaystyle 1=C1+C2-C3+C4 Is this correct so far? 4. Are there any other conditions? 5. Yes but I figured since it only gets messier i'd make sure I got this much correct. The other conditions are: \displaystyle y''(0)=0 and \displaystyle y'''(0)=2 I just do what I did above to find the constants? 6. If what I did above is correct then \displaystyle y''(x)= 2C2e^xcosx-2C1e^xsinx+2C3e^{-x}sinx-2C4e^{-x}cosx then \displaystyle 0=2C2-2C4 \displaystyle y'''(x)= 2C2e^x(cosx-sinx)-2C1e^x(sinx+cosx)+2C3e^{-x}$$\displaystyle (cosx-sinx)-2C4e^{-x}(cosx+sinx)$ and
$\displaystyle 2= 2C2-2C1+2C3-2C4$

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

7. 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.

8. 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.

9. 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.

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

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

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

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

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

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

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

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 $\displaystyle C_{1}=0, C_{2} = \frac{1}{2}, C_{3} =0, C_{4} = \frac{1}{2}$

11. 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:

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

12. 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.

13. 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
$\displaystyle C2=C4$

14. 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 $\displaystyle C_{3}=0$

15. 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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