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Current time:0:00Total duration:8:28

2nd order linear homogeneous differential equations 2

Video transcript

I've spoken a lot about second-order linear homogeneous differential equations in abstract terms and how you know G is a solution then some constant times G is also solution or if G and H are solutions then G + H is also 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 Y 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 you know 1 times its second derivative plus 5 times its first derivative plus 6 times itself is equal to 0 and now let's let's just do a little bit of DIC take a step back and think about what kind of function you know most functions when I take five the function I take its derivative and then I take its second derivative most times I get something completely different right like if I have if I have you know if Y was I don't know if Y was if Y was I'm just trying to think of if Y was x squared then Y prime would be 2x + y prime prime would be 2 and then to add them together you'd say well how would my X terms cancel out I don't you know it so that you get 0 in the end so you know draw back into your brain and think is there some function that you know when I take its first and second derivatives and third and fourth derivatives it essentially became just becomes the same function maybe the the multiple ty is 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 you know maybe e to the X won't work here you could even try it out right if you did e to the X it won't satisfy this this equation right e to the X you know you get e to the X plus 5 e to the X plus 6 e DX if that would not equal to 0 but maybe it's maybe Y 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 back into this and then see if we can actually solve for an R that makes this equation true it if we can we've we found the solution or maybe we found several solutions so let's try it out let's 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 it's always useful to so Y prime is equal to what derivative chain rule derivative of the inside is R and then to 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 the 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 so and I will switch colors so Y the second derivative that's R squared times e to the R X plus 5 times the first derivative so that's 5 r e to the R X plus 6 times our function 6 times e to the R X is equal to 0 and it something might already be surfacing to you as something we can do to this equation to solve for R if all of these terms on the left I'll have an e to the rx so let's let's let's factor that out so this is equal to e to the rx times R squared plus C 5 r 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 in order for this side of the equation to be 0 what do we know can can e to the rx ever equal zero can you ever get something to some exponent and get zero well we'll know so this cannot equal zero so in order for this left-hand side of the equation to be 0 this term this expression right here has to be zero and I'll do that in a different color so we know if we want to solve for R that this R squared plus five R plus 6 that has to be zero and this this is called the characteristic equation this or this the R squared plus five R plus six is called the characteristic equation and it should be obvious to you that now this is no longer calculus this is 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 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 suitable RS that make this equation true this differential equation true and what are those well the first one is y is equal to e to the minus 2x right R is minus 2 and then we can call that Y 1 and then the second solution we found Y 2 is e to the what is this R is minus 3x now my question to you is are these 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 y 1 is a solution we also know that we can multiply Y 1 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 now it's a little bit more general right it's a whole class of functions it's the C doesn't have to just be a 1 it can be any constant and we'll and then when you use your initial values you actually can figure out what that constant is and same for y2 y2 it doesn't have to be one times e to the minus 3x it has to be some any constant and we learned that in the last video that if something is a solution some constant times that is also 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 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 c1 e to the minus 2x plus c2 e 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 and we just tried out e to the RX maybe there's some other a 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 in previous when we did first-order differential equations we only had one constant and that was okay 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 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 what you know at a given 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 you soon