# 2nd order linear homogeneous differential equationsÂ 4

## Video transcript

Let's solve another 2nd order
linear homogeneous differential equation. And this one-- well, I won't
give you the details before I actually write it down. So the differential equation is
4 times the 2nd derivative of y with respect to x, minus
8 times the 1st derivative, plus 3 times the function
times y, is equal to 0. And we have our initial
conditions y of 0 is equal to 2. And we have y prime of
0 is equal to 1/2. Now I could go into the whole
thing y is equal to e to the rx is a solution, substitute
it in, then factor out e to the rx, and have the
characteristic equation. And if you want to see all of
that over again, you might want to watch the previous
video, just to see where that characteristic equation
comes from. But in this video, I'm just
going to show you, literally, how quickly you can do these
type of problems mechanically. So if this is our original
differential equation, the characteristic equation is going
to be-- and I'll do this in a different color-- 4r
squared minus 8r plus 3r is equal to 0. And watch the previous video if
you don't know where this characteristic equation
comes from. But if you want to do these
problems really quick, you just substitute the 2nd
derivatives with an r squared, the first derivatives with an
r, and then the function with-- oh sorry, no. This is supposed to
be a constant-- And then the coefficient on the
original function is just a constant, right? I think you see what I did. 2nd derivative r squared. 1st derivative r. No derivative-- you could say
that's r to the 0, or just 1. But this is our characteristic
equation. And now we can just figure
out its roots. This is not a trivial one for
me to factor so, if it's not trivial, you can just use
the quadratic equation. So we could say the solution
of this is r is equal to negative b-- b is negative 8, so
it's positive 8-- 8 plus or minus the square root
of b squared. So that's 64, minus 4
times a which is 4, times c which is 3. All of that over 2a. 2 times 4 is 8. That equals 8 plus or minus
square root of 64 minus-- what's 16 times 3-- minus 48. All of that over 8. What's 64 minus 48? Let's see, it's 16, right? Right. 10 plus 48 is 58, then
another-- so it's 16. So we have r is equal to 8 plus
or minus the square root of 16, over 8, is equal to
8 plus or minus 4 over 8. That equals 1 plus
or minus 1/2. So the two solutions of this
characteristic equation-- ignore that, let me scratch that
out in black so you know that's not like a 30 or
something-- the two solutions of this characteristic equation
are r is equal to-- well 1 plus 1/2 is equal to
3/2-- and r is equal to 1 minus 1/2, is equal to 1/2. So we know our two r's, and we
know that, from previous experience in the last video,
that y is equal to c times e to the rx is a solution. So the general solution of this
differential equation is y is equal to c1 times e-- let's
use our first r-- e to the 3/2 x, plus c2 times
e to the 1/2 x. This differential equations
problem was literally just a problem in using the
quadratic equation. And once you figure out
the r's you have your general solution. And now we just have to use
our initial conditions. So to know the initial
conditions, we need to know y of x, and we need to
know y prime of x. Let's just do that right now. So what's y prime? y prime of our general solution
is equal to 3/2 times c1 e to the 3/2 x, plus--
derivative of the inside-- 1/2 times c2 e to the 1/2 x. And now let's use our actual
initial conditions. I don't want to lose them-- let
me rewrite them down here so I can scroll down. So we know that y of 0 is equal
to 2, and y prime of 0 is equal to 1/2. Those are our initial
conditions. So let's use that information. So y of 0-- what happens
when you substitute x is equal to 0 here.? You get c1 times e to the 0,
essentially, so that's just 1, plus c2-- well that's just e to
the 0 again, because x is 0-- is equal to-- so this is,
when x is equal to 0, what is y? y is equal to 2. Y of 0 is equal to 2. And then let's use the
second equation. So when we substitute x is equal
to 0 in the derivative-- so when x is 0 we get 3/2 c1--
this goes to 1 again-- plus 1/2 c2-- this is 1 again, e to
the 1/2 half times 0 is e to the 0, which is 1-- is equal
to-- so when x is 0 for the derivative, y is equal to 1/2,
or the derivative is 1/2 at that point, or the slope
is 1/2 at that point. And now we have two equations
and two unknowns, and we could solve it a ton of ways. I think you know how
to solve them. Let's multiply the top
equation-- I don't know-- let's multiply it by 3/2,
and what do we get? We get-- I'll do it in a
different color-- we get 3/2 c1 plus 3/2 c2 is equal to--
what's 3/2 times 2? It's equal to 3. And now, let's subtract-- well,
I don't want to confuse you, so let's just subtract the
bottom from the top, so this cancels out. What's 1/2 minus 3/2? 1/2 minus 1 and 1/2. Well, that's just
minus 1, right? So minus c2 is equal to--
what's 1/2 minus 3? It's minus 2 and 1/2,
or minus 5/2. And so we get c2 is
equal to 5/2. And we can substitute back
in this top equation. c1 plus 5/2 is equal to 2, or c1
is equal to 2, which is the same thing as 4/2, minus 5/2,
which is equal to minus 1/2. And now we can just substitute
c1 and c2 back into our general solution and we have
found the particular solution of this differential equation,
which is y is equal to c1-- c1 is minus 1/2-- minus 1/2 e to
the 3/2 x plus c2-- c2 is 5/2-- plus c2, which is 5/2, e
to the 1/2 x, and we are done. And it might seem
really fancy. We're solving a differential
equation. Our solution has e in it. We're taking derivatives
and we're doing all sorts of things. But really the meat of this
problem was solving a quadratic, which was our
characteristic equation. And watch the previous video
just to see why this characteristic equation works. But it's very easy to come up
with the characteristic equation, right? I think you obviously see that
y prime turns into r squared, y prime turns into r, and then
y just turns into 1, essentially. So you solve a quadratic. And then after doing that,
you just have to take one derivative-- because after
solving the quadratic, you immediately have the general
solution-- then you take its derivative, use your
initial conditions. You have a system of linear
equations which is Algebra I. And then you solve them for the
two constants, c1 and c2, and you end up with your
particular solution. And that's all there is to it. I will see you in
the next video.