Ordinary (Or Partial) Differential Equation Unique Solution

[highcoughdrop]

Homework Statement

(d⁴y)/(dx⁴) = y. Find the Unique solution y = y(x).

Homework Equations

Boundary Conditions: y(0)=0, y'(0) = 2, y''(0) = 0 , y($$\pi$$) = 0

The Attempt at a Solution

I really don't know where to start. I first started off with the guess that y = c1*sin(Ax) + c2*cos(bx). I got some discrepencies that say that c1 was not equal to c1. I also tried getting the characteristic equation, and then seeing if lambda was an eigenvalue, however I"m not completely sure about how to do this.

 

[rock.freak667]

that is the same as


$$\frac{d^4y}{dx^4}-y=0$$


constant coefficients, so try y=e^mx

 

[highcoughdrop]

It doesn't work because y(0) is never going to be zero, unless the coeffcient in front of it is 0, this y=0. However, usually this is a trivial solution, but in this case that doesn't even work because y'(0) is not = 2

 

[Office_Shredder]

Your equation is satisfied by four functions:

e^x, e^-x, cos(x), sin(x)

What other possible solutions can you construct?

 

[highcoughdrop]

Isnt there a systematic way of solving these? or is it just going to be a bunch of guesses?

 

[highcoughdrop]

So far, my only success at this problem has been when y = 2 sin(x). And i believe that's the end of it. I was just looking for some mathematical way to solve the system.

 

[Office_Shredder]

Basically, a differential equation of degree n will have n "linearly independent" solutions that form a basis. What that means is that it has n functions that are basically distinct from each other, and every other solution can be constructed by summing multiples of those solutions. e^x, e^-x, cos(x), sin(x) are all different, so any solution to your differential equation must be of the form

$$y=Ae^x + Be^{-x} + Ccos(x) + Dsin(x)$$ where A,B,C and D are arbitrary numbers

If you use your four initial conditions, you get four equations and four unknowns.

 

[highcoughdrop]

Ahh, thank you. I think I knew this, I suppose I just need some clarification. I do believe that y = 2sinx is the only thing that is left after applying all of the intial conditions. Thank you much. =)

 

[HallsofIvy]

Actually trying what Office Shredder suggested, instead of complaining that you can't gives, with y= e^mx, m⁴e^mx- e^mx= 0 or, dividing through by e^mx which is never 0, m⁴- 1= 0 or m⁴= 1. That has four solutions. What are they?

 

[highcoughdrop]

The four solutions are 1, -1 -i and i.

 

[rock.freak667]

The four solutions are 1, -1 -i and i.

good so putting these together gives the equation Office_Shredder posted as

$$y(x)=Ae^x + Be^{-x} + Ccos(x) + Dsin(x)$$


Now use y(0)=0, y'(0) = 2, y''(0) = 0 , y(π) = 0 to get the values of A,B,C and D

 

[highcoughdrop]

There are 4 equations to solve and they are :
y(0) = 0 ---> A +B + C = 0
y(pi) = 0 ---> Ae^pi + Be^(-pi) - C = 0
y'(0) = 2 - -> A - B + D = 2
y''(0) = 0 ---> A + B-C = 0

 

[rock.freak667]

yes now solve, it shouldn't be too difficult to do it

 

[highcoughdrop]

After all of the mathematical shenanigans, I get A = B = C = 0 and D = 2. Therefore, y = 2sinx. and I never complained. lol