# Homework Help: Memorizing solutions for differential equations

1. Aug 27, 2016

### jdawg

1. The problem statement, all variables and given/known data

Its been a pretty long time since I've taken differential equations and I'm expected to know the solutions to the kinds of DEs below for my fluid mechanics class.

In class my professor worked a 2nd order DE: dy2/dx2 = -k2*y
and told us the way to think about it was to ask yourself "what function gives an alternating solution when derived?" So then the solution he gave us was y = C1*sin(kx)+C2*cos(kx).
I understand that when you integrate/derive a sin or cos function it gives back a sin or cos, but I don't understand how you would know that the function needed to repeat itself... I feel like I didn't really get an explanation as to how you would even know to ask that question to help you memorize the solution. I hope that makes sense.

Another 2nd order DE he worked was: dy2/dx2 = k2*y
and told us the way to think about it was to ask yourself "what function when differentiated twice gives back the same function?" And the solution was:
y = C1*ekx+C2*e-kx. And again, I understand that ex derived/ integrated is ex, but how does thinking this help me memorize the solution?

I also didn't understand this example: dv/dt = -k*v
Solution: v = C1*e-kt

Thanks for any help!

2. Aug 27, 2016

### LCKurtz

3. Aug 27, 2016

### jdawg

THANK YOU!! Exactly what I needed!

4. Aug 27, 2016

### Molar

This is interesting. I never thought it !

For what the professor gave you is a shortcut to remember.
It is said in the equation d2y/dx2 = -k2y itself. You have a function 'y' . Now you differentiate it two times with respect to 'x' and get back 'k' times the original function with the opposite sign. Now which function behaves like this ?
Only if, $$y = sin kθ$$
or, $$y = cos kθ$$
or, $$y = e^{ikx}$$

Again eikx = cos kx + i sin kx.
That's how you know the general solution is this.

Similarly in the next equation, you get the original function back with unaltered sign. Only an exponential function (without 'i ' in it) behaves in that way.

5. Aug 27, 2016

### jdawg

Cool thanks! :)