Differential Equations

Verifying a Solution to a Differential Equation

The final few pages of this class will be devoted to an introduction to differential equation.  A differential equation is an equation (you will see an "=" sign) that has derivatives.  Some examples are

        y' + 3xy = sin x        and        x2y'' + 3xy' + 4y  =  0

They arise when we know of a relationship between how something is changing, how much of that something there is, and how long the process has been going on.  In algebra when we are told to solve, it means get "y" by itself on the left hand side and no "y" terms on the right hand side.  If y = f(x) is a solution to a differential equation, then if we plug "y" into the equation, we get a true statement.  Often, to plug in we need to take derivatives first.


Verify that

        y  =  e3x

is a solution to the differential equation

        y'' + 2y' - 15y  =  0


We compute the first two derivatives.

        y'  =  3e3x         y''  =  9e3x

Now substitute y, y', and y'' into the differential equation.

        9e3x + 2(3e3x ) - 15(e3x )  =  e3x (9 + 6 - 15)

        =  e3x (0)  =  0



Verify that

        y  =  C1ex + C2xex

is a solution to the differential equation

        y'' - 2y' + y  =  0

Particular Solutions

In the exercise above, we saw that the general solution involved constants.  The graph below shows the solution for several choices for the solutions.


Often, we are given an initial value or values that will assist us to find the constants.


The exercise above stated that

        y  =  C1ex + C2xex

is a solution to the differential equation

        y'' - 2y' + y  =  0


        y(0) = 2        and        y'(0)  =  3

find the particular solution



We plug in 0 for t and set y to 2.

        2  =  C1e0 + C2(0)e0 =  C1


        y  =  2ex + C2xex

Now take a derivative to get

        y'  =  2ex + C2(ex + xex)

Plug in 0 for t and 3 for y' to get

        3  =  2e0 + C2(e0 + (0)e0)  =  2 + C2

so that

        C2  =  1

The particular solution is

        y  =  2ex + xex



Find the particular solution to

        y'' + 4y  =  0         y(0)  =  4        y(p/3)  =  1

Given that

        y  =  C1sin(2t) + C2 cos(2t)


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