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## Linear homogeneous equations

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# 2nd order linear homogeneous differential equations 3

## Video transcript

In the last video we had this
second order linear homogeneous differential
equation and we just tried it out the solution y is equal
to e to the rx. And we figured out that if you
try that out, that it works for particular r's. And those r's, we figured out in
the last one, were minus 2 and minus 3. But it came out of factoring
this characteristic equation. And watch the last video if
you forgot how we got that characteristic equation. And we ended up with this
general solution for this differential equation. And you could try it
out if you don't believe me that it works. But what if we don't want the
general solution, we want to find the particular solution? Well then we need initial
conditions. So let's do this differential
equation with some initial conditions. So let's say the initial
conditions are-- we have the solution that we figured
out in the last video. Let me rewrite the differential
equation. So it was the second derivative
plus 5 times the first derivative plus 6 times
the function, is equal to 0. And the initial conditions we're
given is that y of 0 is equal to 2. And the first derivative
at 0, or y prime at 0, is equal to 3. What does y equal at the point
0, and what is the slope at 0-- at x is equal to 0--
and the slope is 3. So how do we use these to
solve for c1 and c2? Well, let's just use the first
initial condition. y of 0 is equal to 2, which is equal
to-- essentially just substitute 0 in into
this equation. So it's c1 times e to the
minus 2 times 0, that's essentially e to the 0,
so that's just 1. So it's c1 times 1, which is
c1, plus c2 times e to the minus 3 times 0. This is e to the 0,
so it's just 1. So plus c2. So the first equation we get
when we substitute our first initial condition is essentially
c1 plus c2 is equal to 2. Now let's apply our second
initial condition that tells us the slope at x
is equal to 0. So y prime is 0. So this is our general solution,
let's take its derivative, and then
we can use this. So y prime of x is
equal to what? The derivative of this is equal
to minus 2 c1 times e to the minus 2x. And what's the derivative
of this? It's minus 3 c2 times
e to the minus 3x. And now we can use our initial
condition, y prime at 0. So when x is equal
to 0, what's the right-hand side equal? It's minus 2 times c1 and then
e to the minus 0, e to the 0, that's just 1. Minus 3 c2, and then once again
x is 0, so e to the minus 3 times 0,
that's just 1. So it's just 1 times
minus 3 c2. And it tells us that when x is
equal to 0, what does this whole derivative equal? Well, it equals 3, right? Y prime of 0 is equal to 3. So now we go back into your
first year of algebra. We have two equations-- two
linear equations with two unknowns-- and we could solve. Let me write them in
a form that you're probably more used to. So the first one is c1 plus
c2 is equal to 2. And the second one is minus 2
c1 minus 3 c2 is equal to 3. So what can we do? Let's multiply this
top equation by 2. There's a ton of ways to solve
this, but if you multiply the top equation times 2, you'll
get-- and I'll do this is a different color, just so that
it's changed-- I'm just multiplying the top one by 2,
you get 2 c1 plus 2 c2 is equal to 4. And now we can add these
two equations. Minus 2 c1 plus 2,
those cancel out. So minus 3 plus 2, you get
minus c2 is equal to 7. Or we could say that c2
is equal to minus 7. And now we can substitute
back in here. We have c1 plus c2-- c2 is minus
7-- so minus 7, is equal to 9, or we know that c--
oh sorry, no I'm already confusing myself. My brain was getting
ahead of myself. c1 plus c2, that's minus 7,
is equal to 2, right? I'm just substituting back
into this differential equation-- sorry, to this
equation, not a differential it's just a simple linear
equation-- and then we get c1 is equal to 9. And now we have our particular
solution to the differential equation. So this was our general
solution. We can just substitute our c1's
and our c2's back in. We have our particular solution
for those initial conditions. And I think that warrants
a different color. So our particular solution is
y of x is equal to c1, which we figured out is 9e to the
minus 2x, plus c2-- well, c2 is minus 7-- minus 7e
to the minus 3x. That is the particular solution
to our original differential equation. And it might be a good exercise
for you to actually test it out. This particular solution to this
differential equation. I'll do another example
in the next video. I'll see you soon.