Stuck on what appears to be simple ODE

[Airsteve0]

I am completely stuck on where to go with the following ODE:

(D^4 + 1)y = 0 where D=d/dx

I know that trying y=e^(rt) is the obvious solution, however, when you solve this you get r^2 = +-i. At this point I am unsure of what to do becuase if I take the square root of "i" I am unsure of how I will represent it in a general solution. Any assistance would be greatly appreciated, thanks!

 

[Matterwave]

An equation like r^4+1=0 means that there are 4 solutions. Instead of putting r^2=+/-i, you can look at r^2=e^(pi/2)*i or e^(3pi/2)*i. Taking the square root, you simply get r=+/-e^(pi/4)*i or +/-e^(3pi/4)*i.

The trick is to put it in the phase explicitly, so that you have an answer.

 

[lurflurf]

(D^4 + 1)=(D^2+sqrt(2)D+1)(D^2-sqrt(2)D+1)
=(D^2+i)(D^2-i)=(D-√ i)(D+√ i)(D-√- i)(D+√- i)
So write a solution in terms of complex exponentials or products of exponentials and trigonometric functions like (e^r x)cos(a x) according to preference.

 

[dimension10]

I am completely stuck on where to go with the following ODE:

(D^4 + 1)y = 0 where D=d/dx

I know that trying y=e^(rt) is the obvious solution, however, when you solve this you get r^2 = +-i. At this point I am unsure of what to do becuase if I take the square root of "i" I am unsure of how I will represent it in a general solution. Any assistance would be greatly appreciated, thanks!

$$\frac{{\mbox{d}}^{4}y}{{\mbox{d}}^{4}x}+y=0$$

$$y={e}^{rx}$$

$${r}^{4}+1=0$$

$${r}^{2}=\pm i$$

$$r=\pm\frac{1\pm i}{\sqrt{2}}$$

$$y={C}_{1}\exp \left(\frac{1+i}{\sqrt{2}}\right)+{C}_{2}\exp \left(\frac{1-i}{\sqrt{2}}\right)+{C}_{3}\exp \left(-\frac{1+i}{\sqrt{2}}\right)+{C}_{4}\exp \left(\frac{i-1}{\sqrt{2}}\right)$$

We can solve for the specific solution for y if and only if we know y(0), y'(0), y''(0), y'''(0) and y''''(0)

 

[Airsteve0]

thanks everyone, I greatly appreciate the help!

 

[dimension10]

thanks everyone, I greatly appreciate the help!

Just realized I made a LaTeX error. Should be

$$y={C}_{1}\exp \left(\frac{1+i}{\sqrt{2}}x\right)+{C}_{2}\exp \left(\frac{1-i}{\sqrt{2}}x\right)+{C}_{3}\exp \left(-\frac{1+i}{\sqrt{2}}x\right)+{C}_{4}\exp \left(\frac{i-1}{\sqrt{2}}x\right)$$

and you may want to use euler's formula to put in some trigonometric functions there.