A differential equation states how a rate of change (a "differential") in one variable is related to other variables.For example, the single spring simulation has two variables: time t and the amount of stretch in the spring, x. If we set x = 0 to be the position of the block when the spring is unstretched, then x represents both the position of the block and the stretch in the spring. Velocity is (as usual) the time derivative of position v = x', and the differential equation describing the single spring simulation is
v' = −k x
where k is the spring constant (how stiff the spring is). Now we can "read" the meaning of the differential equation: it says thatthe rate of change in velocity is proportional to the positionFor instance, when the position is zero (ie. the spring is neither stretched nor compressed) then the velocity is not changing. This makes sense, because the spring is not exerting a force at that moment.
x^{2} + 2x + 1 = 0
which has a solution x = −1 For a differential equation, the solution is not a single value, but a function. The task is to find a function whose various derivatives fit the differential equation over a long span of time. For example,x'' + 2x' + x = 0 | (1) |
x(t) = a e^{−t} + b t e^{−t} | (2) |
x'(t) = (b − a)e^{−t} − b t e^{−t} | (3) |
x''(t) = (a − 2b)e^{−t} + b t e^{−t} | (4) |
x'' + 2x' + x =
=((a − 2b)e^{−t} + b t e^{−t}) + 2((b − a)e^{−t} − b t e^{−t}) + (a e^{−t} + b t e^{−t})
=(a − 2b+2b − 2a+a)e^{−t} + (b − 2b+b) t e^{−t}
= 0
Therefore the solution (2) satisfies the differential equation (1) for any values of a, b.x(0) = 1 x'(0) = 0
Then we can plug t = 0 into equations (2) and (3) above to find the values of the constants a, bx(0) = a + 0 = 1
x'(0) = (b − a) − 0 = 0
and therefore we find that a = b = 1 and the particular solution isx(t) = e^{−t} + t e^{−t}
This solution is called the particular solution because it applies only to the particular initial conditions that we have chosen.This web page was first published June 2001.