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2nd order linear homogeneous differential equations 2

Let's find the general solution! Created by Sal Khan.

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  • leaf green style avatar for user RagnarG
    At , Sal suggests the function e^(rx) to satisfy the equation since it keeps the derivatives "in line", so to speak. That all makes sense to me, but why not generalize it even further to the set of functions e^(rx+k), where k is some arbitrary constant? Wouldn't that be equally valid while covering even more solutions?

    Consequentially the general solution to the diff equation would be y(x) = C_1e^(r_1x + k_1) + C_2e^(r_2x + k_2). Wouldn't that work equally well while covering more answers?
    (5 votes)
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    • leaf green style avatar for user RagnarG
      Nevermind, I realized that the more complicated version doesn't add any solutions since e^(rx + k) = e^(rx) * e^k. The e^k-factor is constant, so the term C_1*e^(rx) already covers the span of C_2*e^(rx + k) - you can just pick the constant C_1 to equal C_2*e^k, and the terms are equal!
      (22 votes)
  • blobby green style avatar for user Carolyn Labutta
    So how do you solve for r if instead of 5 and 6? You just have p and q? y"+py'+qy=0.
    (6 votes)
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  • male robot hal style avatar for user richardpmcnair
    Are the solutions to ALL 2nd order homogenous equation y=e^x functions?
    (6 votes)
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  • blobby green style avatar for user Fabricio Zelada Rivas
    How can we proof these are the only solutions? meaning where I can find a method or way to decide these are the only solutions to these problem.
    (5 votes)
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  • leafers seed style avatar for user Zulfidin Hojaev
    at , why sal multiplied c's to the answers, would it not bean answer withoat c's ?
    (2 votes)
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  • male robot hal style avatar for user Brayden
    If instead of that equation equaling 0 it equaled a constant such as 18, would the function still be homogeneous. as it could still be written as y''+5y'+6y-18=0 or is my reasoning incorrect.
    (3 votes)
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    • blobby green style avatar for user yaga410
      Equation y''+5y'+6y=18 is not homogenous. I believe it can be sold by method of undetermined coefficients (presented further in differential equations course). Shortly, the result of equation should be threated like 18+0, so the general solution would be general solution to this equation =0 plus the particular solution to the same equation =18
      (3 votes)
  • blobby green style avatar for user bernat.pla7
    How do we know there aren't other solutions, apart from the exponential one? I trust Salman Khan, but I would like to know the proof. :)
    (3 votes)
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    • leaf green style avatar for user simontse.ubc
      for 1st order, like y'=ky, the proof is quite easy: let y(t) be any solution (perhaps not the exponential one?), then consider the function u(t) = e^(-kt) y(t)

      check that u'(t) = -k e^(-kt) y(t) + e^(-kt) y'(t) by product rule, now since y‘(t) = k y(t), so u'(t) = (-k + k) e^(-kt) = 0!, what can u(t) be? u(t) = C, a constant, so the unknown solution y(t) has the property that e^(-kt) y(t) = C, so y(t) = C e^(-kt) is certain! the proof based on the fact that u'(t) = 0 implies u(t)=C, this is a basic calculus fact, that can further be proved using even more elementary real analysis techniques, however, if you accept this, then the proof for first order linear equations goes like what I've shown.
      (3 votes)
  • piceratops tree style avatar for user Pallav Swarnkar
    Sal is giving the solution of second order differential equation through the method of finding complimentry function (C.F.) or complete solution.
    But can anyone tell me how we get to the result of exponential function?
    (2 votes)
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  • male robot johnny style avatar for user PJ1999
    I don't understand how Sal applied the chain rule to this function. Can someone please explain it to me step-by-step?
    (2 votes)
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    • aqualine ultimate style avatar for user Kyler Kathan
      The only time he used the chain rule was after substituting y = e^(rx). The chain rule states that given:
      y = f(g(x))
      y′ = g′(x) * f′(g(x))
      Substituting the equation in the problem:
      y = e^(rx)
      f(x) = e^x
      g(x) = rx
      f′(x) = e^x
      g′(x) = r
      y′ = r*e^(rx)
      In the video Sal then takes the second derivative using the same process:
      h(x) = r*e^x
      h′(x) = r*e^x
      k(x) = rx
      k′(x) = r
      y′ = h(k(x))
      y′′ = k′(x) * h′(k(x))
      y′′ = r*r*e^(rx)
      y′′ = r²e^(rx)
      (4 votes)
  • blobby green style avatar for user Varun Sivashankar
    At , Sal mentions that there is a proof for why this is the only general solution to a homogeneous first order differential equation. Can someone please provide the proof?
    (2 votes)
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Video transcript

I've spoken a lot about second order linear homogeneous differential equations in abstract terms, and how if g is a solution, then some constant times g is also a solution. Or if g and h are solutions, then g plus h is also a solution. Let's actually do problems, because I think that will actually help you learn, as opposed to help you get confused. So let's say I have this differential equation, the second derivative of y, with respect to x, plus 5 times the first derivative of y, with respect to x, plus 6 times y is equal to 0. So we need to find a y where 1 times its second derivative, plus 5 times its first derivative, plus 6 times itself, is equal to 0. And now, let's just do a little bit of-- take a step back and think about what kind of function-- if I have the function and I take its derivative and then I take its second derivative, most times I get something completely different. Like, if y was x squared, then y prime would be 2x, and y prime prime would be 2. And then to add them together you'd say, well, how would my x terms cancel out so that you get 0 in the end? So draw back into your brain and think, is there some function that when I take its first and second derivatives, and third and fourth derivatives, it essentially becomes the same function? Maybe the constant in front of the function changes as I take the derivative. And if you've listened to a lot of my videos, you'd realize that it probably is what I consider to be the most amazing function in mathematics. And that is the function e to the x. And in particular, maybe e to the x won't work here-- or you can even try it out, right? If you did e to the x, it won't satisfy this equation, right? You would get e to the x, plus 5e to the x, plus 6e to the x. That would not equal to 0. But maybe y is equal to e to some constant r, times x. Let's just make the assumption that y is equal to some constant r times x, substitute it back into this, and then see if we can actually solve for an r that makes this equation true. And if we can, we've found the solution, or maybe we've found several solutions. So let's try it out. Let's try y is equal to e to the rx into this differential equation. So what is the first derivative of it, first of all. So y soon. prime is equal to what? Derivative chain rule. Derivative of the inside is r. And then derivative of the outside is still just e to the rx. And what's the second derivative? y prime prime is equal to derivative-- r is just a constant-- so derivative of the inside is r, times r on the outside, that's r squared, times e to the rx. And now we're ready to substitute back in. And I will switch colors. So the second derivative, that's r squared times e to the rx, plus 5 times the first derivative, so that's 5re to the rx, plus 6 times our function-- 6 times e to the rx is equal to 0. And something might already be surfacing to you as something we can do this equation to solve for r. All of these terms on the left all have an e to the rx, so let's factor that out. So this is equal to e to the rx times r squared, plus 5r, plus 6 is equal to 0. And our goal, remember, was to solve for the r, or the r's, that will make this true. And in order for this side of the equation to be 0, what do we know? Can e to the rx ever equal 0? Can you ever get something to some exponent and get 0? Well, no. So this cannot equal 0. So in order for this left-hand side of the equation to be 0, this term, this expression right here, has to be 0. And I'll do that in a different color. So we know, if we want to solve for r, that this, r squared plus 5r, plus 6, that has to be 0. And this is called the characteristic equation. This, the r squared plus 5r, plus 6, is called the characteristic equation. And it should be obvious to you that now this is no longer calculus. This is just factoring a quadratic. And this one actually is fairly straightforward to factor. So what is this? This is r plus 2, times r plus 3 is equal to 0. And so the solutions of the characteristic equation-- or actually, the solutions to this original equation-- are r is equal to negative 2 and r is equal to minus 3. So you say, hey, we found two solutions, because we found two you suitable r's that make this differential equation true. And what are those? Well, the first one is y is equal to e to the minus 2x, right? We can call that y1. And then the second solution we found, y2 is e to the-- what is this?-- r is minus 3x. Now my question to you is, is this the most general solution? Well, in the last video, in kind of our introductory video, we learned that a constant times a solution is still a solution. So, if y1 is a solution, we also know that we can multiply y1 times any constant. So let's do that. Let's multiply it by c1. That's a c1 there. This is also going to be a solution. And now it's a little bit more general, right? It's a whole class of functions. The c doesn't have to just be 1, it can be any constant. And then when you use your initial values, you actually can figure out what that constant is. And same for y2. y2 doesn't have to be 1 times e to the minus 3x, it has to be any constant. And we learned that in the last video, that if something's a solution, some constant times that is also a solution. And we also learned that if we have two different solutions, that if you add them together, you also get a solution. So the most general solution to this differential equation is y-- we could say y of x, just to hit it home that this is definitely a function of x-- y of x is equal to c1e to the minus 2x, plus c2e to the minus 3x. And this is the general solution of this differential equation. And I won't prove it because the proof is fairly involved. I mean, we just tried out e to the rx. Maybe there's some other wacko function that would have worked here. But I'll tell you now, and you kind of have to take it as a leap of faith, that this is the only general solution. There isn't some crazy outside function there that would have also worked. And so the other question that might be popping in your brain is, Sal, when we did first order differential equations, we only had one constant. And that was OK, because we had one set of initial conditions and we solved for our constants. But here, I have two constants. So if I wanted a particular solution, how can I solve for two variables if I'm only given one initial condition? And if that's what you actually thought, your intuition would be correct. You actually need two initial conditions to solve this differential equation. You would need to know, at a given value of x, what y is equal to. And, maybe at a given value of x, what the first derivative is. And that is what we will do in the next video. See