### 載入中...

相關課程

⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.

# 2nd Order Linear Homogeneous Differential Equations 4 : Another example with initial conditions!

相關課程

選項
分享

0 / 750

- 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.

載入中...