How to Solve a Second-Order Differential Equation with Constants h and j?

[youngoldman]

d²y/ dx² = - h y / j

where h, j are constants. What's y?

 

[youngoldman]

This on its own is not actually a question in my assignment but it is a starting point I need to get the problem done.

 

[gabbagabbahey]

Can you think of a function whose second derivative is proportional to it?

 

[youngoldman]

I know y is an exponential function with a multiplication constant out the front, just not sure what the argument of the exponential is.

 

[youngoldman]

Actually I know it's going to be the sum of two exponentials with different multipliation contants and one will have the negative argument of the other.

 

[youngoldman]

In other words I know it's in the form

y = Aexp(c) + B exp (-c), just not sure what the c is.

 

[gabbagabbahey]

Since, the derivative is respect to x, why not try kx where k is a constant (it may be complex) for your argument? In fact, the general solution has two terms y(x)=Ae^{kx}+Be^{-kx}. When you plug this into your DE, what do you get?

 

[youngoldman]

and k = sqrt (- h/j)?

so it is a complex argument as I was expecting because a complex exp can be written in terms of sines and cosines, whose 2nd derivative is their own negative.

 

[gabbagabbahey]

Yes, exactly so you may as well write y(x)=Csin(\frac{h}{j}x)+Dcos(\frac{h}{j}x)

 

[youngoldman]

Thank you, gabbagabbahey.

 

[klondike]

This is a ay''+by'+cy=0 problem which has been discussed to death on every DE book. One should be able to write down the results (when b^2-4ac>0,<0 and = 0) while sleeping.