LSO -Two different real roots

Up a level : Differential Equations
Previous page : Linear second order equations with constant coefficients
Next page : LSO - Two equal real roots

In this case, we get as in our example. The general solution is

y = C{e^{{r_1}x}} + D{e^{{r_2}x}}

where r1 and r2 are the roots of the characteristic equation.  We cannot make a slope field though, because we can have several solutions going through one point. Let us look at the example from the previous page.

y'' + 2y' - 15y = 0

with the solution

y = C{e^{3x}} + D{e^{ - 5x}}

Let us say we want our solutions to go through the point (0, 1). We thus get that

1 = C{e^0} + D{e^0} = C + D

So D=1-C. The solutions will now be

y = C{e^{3x}} + (1 - C){e^{ - 5x}}

This is plotted for various values of C down below.

Up a level : Differential Equations
Previous page : Linear second order equations with constant coefficients
Next page : LSO - Two equal real rootsLast modified: Dec 28, 2023 @ 17:33