Memorizing solutions for differential equations

[jdawg]

Homework Statement

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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*e^kx+C₂*e^-kx. And again, I understand that e^x derived/ integrated is e^x, 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^-ktThanks for any help!

 

[LCKurtz]

Your examples are all constant coefficient DE's. They are easy to solve and there is much information on the internet. One example is
https://www.cliffsnotes.com/study-g.../second-order-equations/constant-coefficients
Take a look at that.

 

[jdawg]

THANK YOU! Exactly what I needed!

 

[Molar]

This is interesting. I never thought it !

For what the professor gave you is a shortcut to remember.

but I don't understand how you would know that the function needed to repeat itself

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 e^ikx = 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.

 

[jdawg]

Cool thanks! :)