Real and distinct roots, Mathematics

Now we start solving constant linear, coefficient and second order differential and homogeneous equations. Thus, let's recap how we do this from the previous section. We start along with the differential equation.

ay′′ + by′ + cy = 0

 Write down the feature equation.

 ar2 + br + c = 0

 So solve the characteristic equation for the two roots r1 and r2. It provides the two solutions

y1(t) = er1t and y2(t) = er2t

Here, if the two roots are real and distinct that is "nice enough" by the general solution  r1 ≠ r2. This will turn out that these two solutions are as

 y (t )= c er1 t + c er2 t

As with the previous section, we'll ask that you believe us while we means that such are "nice enough". You will be capable to prove this simply enough once we reach a later section.

With real, distinct roots there actually isn't an entire lot to do other than work a couple of illustrations so let's do that.

Posted Date: 4/10/2013 5:10:32 AM | Location : United States

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