The most important fact about equations involving numbers is -- You can always check a possible answer. For example, if you think that x = 7 might be the solution of the equation
all you need to do is try it out
2 x + 3 ?=? 17
2 (7) + 3 ?=? 17
14 + 3 ?=? 17
17 ?=? 17 -- YES!!
So x = 7 is a solution of the equation 2 x + 3 = 17.
The same idea works for differential equations also.
Example:
Is y = sin t a solution of the differential equation y'' + y = 0?
y'' + y ?=? 0
(sin t)'' + (sin t) ?=? 0
-sin t + sin t ?=? 0
0 ?=? 0 -- YES!!
So y = sin t is a solution of this differential equation.
is a solution of the equation y' + 5y = 0. answer
is a solution of the equation y'' + 5y' + 4y = 0. answer
One of the most powerful methods of solving differential equations is called Guess-and-Check. You can use any method that you want to figure out the solution to a differential equation and then determine whether it is right by checking it as we did above. This method is particularly powerful when used with computers or calculators and graphics. Your CAS window will describe how to use your computer algebra system to produce the graph of a solution of a differential equation. Based on such a graph you can often guess the solution and then you can check your guess.
Example:
The Mathematica session below shows the solution of the initial value problem
y'' = -3y, y(0) = 0, y'(0) = 1
Looking at this graph it looks like a modification of the sine function. The amplitude is not 1 and the period is not 2 pi, so we might guess that this particular solution is of the form
We can check our guess and determine the values of the constants A and w as follows.
y = A sin (w t)
y' = Aw cos (w t)
y'' = -Aw2 sin (w t)
y'' ?=? -3 y
-A w2 sin (w t) ?=? -3 A sin w t)
w2 ?=? 3
and we see that our guess works if w is the square root of three. Notice that the amplitude A doesn't matter as long as we are looking for any solution of the differential equation. This differential equation has many different solutions and the graph above shows just one of them.
Now we bring the initial conditions into the picture. Since sin 0 = 0 the first initial condition will be satisfied no matter what A is. Next we look at the second initial value.
y'(0) ?=? 1
Aw cos (w 0) ?=? 1
Aw ?=? 1
1
A ?=? ---
w
and we see that a must be 1/w.
Notice this is the same differential equation as the example above but that the initial conditions are different from the ones above. Guess what the solution to this IVP is and check your guess as in the example above.
Then guess what the solution is and check your guess as in the example above.
Then guess what the solution is and check your guess as in the example above.
Then guess what the solution is and check your guess as in the example above.