A differential equation like y'' + y = o is important because it appears in physics. I am confused. How does it appear in physics?
I presume, since you are asking about differential equations, that you have taken calculus and know that y'' is the second derivative. If, for example, y is the position of some object, say its distance from a given point, then y', the first derivative, is the rate of change of position- the speed at which that object is moving. And y'', the second derivative, is the rate at which that speed is changing- the acceleration of the object. Second order differential equations are especially important in physics because "Force equals mass times acceleration". Taking mass to be 1, y''+ y= 0, y''= -y, means that the acceleration, and so the force is the negative of the distance- a
spring is an example of such a force- the more it is stretched the more the force is back the other way.
Because doesn't y stand for f(x) which can be any function? like x^2 + 1
Well y can be any function that
satisfies that equation! No y cannot be \(\displaystyle x^2+ 1\) because if \(\displaystyle y= x^2+ 1\), then \(\displaystyle y'= 2x\) and \(\displaystyle y''= 2\). \(\displaystyle y''+ y= 2+ x^2+ 1\) is not equal to 0.
So, without knowing the actual function in terms of x, how does solving y'' + y = 0 help? I know this is probably a bad question. But I am a newbie. Do you plug in f(x) after you solve for y?
I am tempted to say that you
start by taking a Calculus class!
One of the things that you learn in a Calculus class is to take derivatives and, in particular, that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x). So if y= sin(x) then y''= -sin(x)= -y. Similarly if y= cos(x) then y''= -cos(x)= -y. We also know, from the general theory of linear equations, that any linear combination of solutions is also a solution: The general solution to this differential equation is y(x)= Acos(x)+ Bsin(x).