LSO -Two different real roots

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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.

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Last modified: Dec 28, 2023 @ 17:33