7 6 7 7 Finding Solutions using Separation of Variables

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[00:00:09]okay welcome back everybody uh today we are going to discuss finding solutions using a separation of variables um i've discussed this in unit six so um this should be somewhat of a review for you which is a good thing so let's go ahead and get started it says up till now we've only been able to solve two different types of equations when you're getting a first derivative is equal to some function in terms of x or when you're given a second derivative of is equal to some function in terms of x but what about when that function is uh or that yeah function i suppose is in terms of x and y so something like this the first derivative is equal to 2x divided by y so when you have an equation with two variables the first thing that's most important sometimes you'll see this on a free response is the separation of variables probably one of the most important parts so anything with x's on one side anything with y's on the other side now this first derivative we can also be rewrite that as um d y d x okay that's equivalent to that um now there if you recall i brought up the other form which would be d y equals 2x divided by y dx you recall that form that we've discussed in unit 6.

[00:01:53]leibniz um that was his form um and technically it's not multiplying both sides by delta x it's just two different forms of the same equation but if you want to think of it as multiplying it's not going to be wrong if you think of it that way but technically that's not what we're doing um so this little y here is is uh not supposed to be with the x's so we're going to move that over here which we are going to be multiplying both sides by y so you have y d y equals 2x dx so now that is a separation of variables and that usually gets you a point if you see it on a free response okay so what after you separate the variables you can take the integral of both sides and this side you're going to take the integral of y with respect to y and this you're going to take the integral of x with respect to that dx or x so again for this one it's just power rule let's go ahead and integrate with respect to y we're going to add to the exponent and divide okay and it should be plus c but if you again if you recall we only need a one plus c on one side because the c's um combine into one c so over here again we can keep that two that coefficient out in front so we are adding to the exponent and dividing by our new exponent and then don't forget the plus c okay so the twos end up canceling out um we're going to multiply both sides by 2 and you get y squared equals 2 x squared plus 2 times a constant is still some constant and then to solve for y you take the square root of both sides and do not don't forget your plus and your minus okay and that's your answer okay so that's a general solution because obviously we don't know enough information to find c but that is finding a general solution uh using separation of variables so here's a couple examples of how to separate variables again we would subtract the x squared and move the dx to the other side this one we would divide by sine x and again this is d y over dx this one right here let's move the x to the other side okay and then again separate the d y on one side and the dx on the other so there's a couple of rewritten with separation of variables so let's go ahead and try this one this one has the y so move the y over here again the opposite of dividing to undo dividing is multiplying so let's go ahead and try this one we're going to write that as the cosine of y d y equals 4 x cubed dx okay and then again once you have the separation of variables the next step is to integrate both sides what's the antiderivative of cosine y antiderivative of cosine that is positive sine again we've got our coefficient here we're going to add to the exponent and divide and then plus c okay so again the fours will cancel and again you're trying to solve for y when it asks you to solve the differential equation they want you to solve for y so how do we get rid of sine that's arc sine and that is your solution okay so there's two examples general solutions using separation of variables so nothing nothing too complicated now this is finding a particular solution and the reason why we can find a particular solution is because they give us an initial condition okay and there's two ways to go about doing this i'm going to start with the way that most of you have been solving your um like when you're integrating when you're turning velocity into position and that type of thing you were given an initial condition so this is how most of you were doing it first thing is um [Music] let me change colors here first thing is to write it in liveness notation which is this notation here okay well that's also a separation of variables so we separated the variables now what gets some people is the integration of the left side you're like well how do i integrate d y well you're integrating one is basically what you're doing you're integrating one d y you're integrating one with respect to y so what would that be that would just be y now on this side you're integrating x to the one third so integrating x to the one third you're going to add one right one to the exponent which gives you four four thirds and then divide by four thirds which is the same as multiplying by three fourths and don't forget your plus c now the initial condition is when x is equal to 1 y is equal to 2. so we can go ahead and we can substitute that information in 2 equals 3 4 times 1 to the 4 3 power now this is something you have to remember how to do without your calculator this is the cube root of 1 to the 4th power and again it's no big deal because that's just a 1 1 raised to anything is 1 but if this was something other than 1 cube root fourth power so we end up getting oops i'm sorry i forgot my plus c so all of this is equal to one times three fourths so c equals three-fourths i'm sorry two equals three-fourths plus c so what does c equal to well this is fourths minus three-fourths so c is five-fourths now a lot of people stop there if you're like oh i solved no you solve for c they want you to find the particular solution so again you need to understand what this means particular solution it means whose derivative gives you this and this point works in that so that solution is y equals three fourths x to the four thirds plus the c value you calculated this is the solution this is the particular solution whoops so let's go ahead and plug in one into here if you plug in one into here you end up getting three-fourths plus 5 4 which is 8 4. do you get the result of 2 this has to work in your solution so if there's a plus or minus option you have to see which one is it is it plus or is it minus something to think about now this is how most of you have been solving these um there is another way i showed you which is the way that i tend to do it and that is using your initial condition so if you go to take this original here and you integrate d y but you use your initial condition 2 plus 1 to the x and i'm just going to change that to x to the one third oops that should be we're going to use our dummy variable t okay that's just our dummy variable so they're not all x's remember when we did that this is the initial condition so this is another way to solve and you can do this with the acceleration the velocity from velocity to position this is the the given right here okay and this is the way i tend to do them but again it's personal preference so when i integrate d y i get y and over here when i integrate this i end up getting t to the four thirds again multiplied by three fourths and i'm evaluating that from one to x that's how it's going to be different so i'm gonna go ahead and i do y equals two plus i plug in my x i get three fourths x to the four thirds there's my f of b minus my f of a three-fourths times one to the four thirds and again you could keep that three-fourths out in front and multiply it in after but i'm just going to use it as one whole term here so there's my f of b minus my f of a okay so i get the same answer 3 4 x to the four-thirds this is right here um a one so that's minus three-fourths okay so i get two minus three-fourths i get eight-fourths minus three-fourths which is five-fourths same answer okay so this is integrating and plugging your c in afterwards which again from what i'm seeing most of you are doing this this is using the initial condition um what do we call this the net change theorem i believe if you want to look back in your notes okay again personal preference this is my preference i'm not saying one is right one is right one is wrong i'm not saying that okay so again if you see these on a free response which it is possible more so on the a b test but i do see it on the bc test these are the steps that you need to take okay separate the variables take the antiderivatives of both sides use the initial condition to find the constant of integration that's our plus c solve the equation for y and then sometimes you have to use that equation to evaluate something okay so all of this it's it's nice if they give you one of these because this is usually worth four or five points out of nine i will tell you that solving for y sometimes the math on solving for y is quite crazy and it might not be something you want to waste your time doing just for one point okay so let's go ahead and look at this one given the differential equation find the particular solution there's that those words again particular solution determined by the initial condition okay so again i'm not sure which which method um you like better um but i'm going to just stick with mine if you want to do you want to stop the video and go ahead and try it's probably a good idea okay so step one separate the variables so that's y d y now be careful here this is x plus one dx okay step two is to integrate both sides and again you can do that right within step one you don't have to rewrite it but i'm going to rewrite it anyway okay so we're integrating y with respect to y so that's the power rule and here we're going to re uh power rule on this one and plus c okay it says use the initial condition to find the constant of integration i'm just going to go with this because again i guess most of you do it this way anyway so let's just go ahead and go with this we'll go with your method um [Music] all right what are we given we are given when x is equal to 0 y is negative two that's what we're going to plug in into here so negative two squared divided by two zero squared divided by two plus zero plus c so that is two equals c okay so we found our constant of integration um and we're going to plug it back into step 2 y squared divided by 2 equals x squared divided by two plus x plus two now again this sometimes this requires um logarithmic um all your log rules and it can be kind of hairy to solve for y and it's only worth one point but this isn't too bad um so we'll go ahead and solve this one so multiply through by two and you get x squared plus two x plus four take the square root of one of both sides you get plus or minus the square root of x squared plus two x plus four okay so again remember that your solution has to work this has to work in your solution so we have to see which one as i was saying before which one is it the plus or the minus so i'm going to go back up here and when i plug 0 in for my y i'm sorry zero in for my x do i get the plus or do i get the minus um plug negative 2 in for my y sorry you can't see that negative 2 in for my y 0 in for my x this has to work in my solution so when x is equal to zero here we have zero squared plus two times zero i think it's pretty obvious which one it's going to be right plus or minus the square root of four is it plus or minus the square root of four that we want well obviously we want the negative 2.

[00:18:05]so our solution is y equals negative square root of x squared plus 2x plus 4. so if you forget your plus or minus here in your work and you just automatically pick plus if it were to be a plus you're not going to get credit because they want you to test that point especially when you take the square root of both sides it will be plus or minus you then choose is it plus or is it minus so this is the solution and if you want to check if you take the derivative of this and you replace this y into here they will be equal to each other but you don't have to check for that and then five use this equation to evaluate a value well they didn't ask us that so we're done okay all right so let's try another one again if you want to go ahead and stop the video and try it on your own might be a pretty good idea so the first thing we're going to do is separate the variables so i'm going to get the y's on one side so that's actually 1 over y on this side and then here it's going to be x divided by x squared plus 4.

[00:19:39]okay we're going to go ahead and integrate both sides and again you can just do this right in here you just do the integral that's like two steps in one which is completely fine all right how do you integrate one over y with respect to y that's the natural log of the absolute value of y and again plus c but we're only going to do it on one side because plus c and a c is still a c all right how do we integrate this well this is a rational right a rational we're integrating a rational function that tends to be u substitution let me set my u equal to my denominator here which is usually the first thing we try so d u equals 2x dx here's my x dx right there okay so it's going to be equal to the integral my x it's going to be 1 over u and then what is x dx well x dx is d u divided by a 2.

[00:21:02]all right so the natural log of the absolute value of y is equal to one half times natural log of our u x squared plus four plus c and we're gonna plug in the initial condition when x is negative three y is equal to two and find our c value okay so the natural log of the absolute value of two equals one half the natural log of our x squared plus four the natural log of two equals one half the natural log of what is this thirteen trying to solve for c how might one solve for c well c is equal to the natural log of two absolute value of 2 is 2 minus 1 half the natural log of 13.

[00:22:22]that's our c value now we can condense that and if it's multiple choice i'm going to tell you that that's not going to be you'll see this in the multiple choice if you don't see it in free response so let's go back up to our step um two right here yeah two because we're working on it so we're gonna go back up here and we're gonna take oops the natural log of the absolute value of y is equal to one half the natural value or the natural log of the absolute value of x squared plus four plus our c value okay so on a free response you found the particular solution but you haven't solved it for y and again this is something you want to do last solving all of this for y okay many different ways you can solve for y it's y in terms of x um all of this work i'm about to do is only worth one point so again you might want to do it at at the end um first of all i'm going to take my exponents up here so i can use my log properties remember that and i know how much you love your logarithms so by using my log properties and condensing this of the sum and the difference i get the natural log of x squared plus four i don't need the absolute value because that is positive um raised to the one-half that's this one right here and because this is a plus i multiply those together so i'm just gonna do a times two and because this is a subtraction i divide by 13 to the one-half which is the square root of 13 or however you want to write it 13 to the one-half or the square root of 13.

[00:24:32]okay so we have a natural log equals natural log so you can use the equality property see natural log equals a natural log so the equality property means that these two things have to be equal to each other so the absolute value of y has to be equal to all of this i'm just going to write it like this this is the same thing right so if the absolute value of y equals this then y either equals plus or minus all of that and again which one is it well we have to test negative three and two if i plug in a negative 3 in for x which one's going to give me a positive 2.

[00:26:01]is it going to be a plus or the minus so 2 square roots of 13.

[00:26:10]i plug in my negative 3 squared so that's going to give me 13 to the one-half which again is the square root of 13.

[00:26:21]so these cancel i need my result for y to be a positive two so i am going to choose oops wrong one i'm going to choose the positive so your solution for this is y equals positive 2 over the square root of 13.

[00:26:52]times x squared plus 4 to the one-half and you can write that as a square root if you prefer it really doesn't make a difference okay now that one wasn't too bad if you remember your log properties um sometimes only one side has the natural log and you have to remember how to change it from log form to exponential form quickly i'll review that if you have like the natural log of y equals i don't know um two to the x plus three i'm just making something up then y equals e to the two x plus three if you remember that changing from log form to exponential form this is your base a log is equal to the exponent which equals that so this again remembering that property from from 10th grade if you don't again you're just going to lose a point not that big of a deal okay one more example again if you want to stop and go ahead and try this one on your own um that would be a good time to do that okay so separation of variables would be one over y d y equals one over x squared dx okay go ahead and take the integral of both sides we must do a separation of variables first otherwise you lose all the points no matter what it's the first thing that has to be done okay so here we have the natural log of the absolute value of y again and how do we integrate 1 over x squared you guys remember that something whose derivative is 1 over x squared well it's negative 1 over x because the derivative of one over x is one over x squared negative one over x squared so the derivative of negative one over x is positive 1 over x squared so try to remember that one because that one usually will come up on the test or if you really want to you can change this to a negative exponent and do it the long way but i would just memorize that again and do it in your head okay so the next thing we have to do is find our constant of integration what are we given we're given one three so the absolute value of our y value negative 1 divided by our x value the natural log of 3 equals negative 1 plus c so that tells us that c is equal to the natural log the absolute value of 3 is 3 plus 1. so be careful with this if you want to keep the absolute value around it or put parentheses around it you can because i don't want to adding the 3 and the 1 together okay so we go back up to our integration the natural log of the absolute value of y equals negative 1 over x plus your c value okay now we need to solve for y so notice they're not all natural logs so we cannot combine all of these we need to try to get the y by itself and this is where that property comes in that i was just telling you about a log is equal to an exponent so all of this is your exponent so the absolute value of y is equal to e to the negative 1 over x plus the natural log of the absolute value of 3 plus 1. that whole thing is your exponent okay all right so y is equal to plus or minus all of this now again i'm going to give you a hint for when these are multiple choice i mean could you write plus or minus all of this and then test your test point absolutely but i'm going to show you if this is multiple choice it's not going to look like this your exponent you have to remember some algebra 1 properties if i take e to the negative 1 divided by x times e to the natural log of 3 times e to the first power isn't it true that's an x right there and it doesn't look like it so let me fix that isn't it true that when the bases are the same you can add the exponents so when you're adding the exponents you can split it up into a product with the same base something you might have to do to get one of the multiple choice answers so i'm gonna refreshing your memory with this is the hardest part of this chapter or this unit is all the log properties that you may have you may have to use okay so and this should be plus or minus because it's absolute value okay so we have e to the negative 1 over x which is just e to the negative 1 over x what is e to the natural log of 3 well again that's just three and what is e to the first that is also e so we have three times e to the first that this is acceptable this is acceptable or y equals plus or minus 3 e to the negative 1 over x plus one that's acceptable but again you have to determine is it plus or is it minus any of these forms are acceptable but you have to determine plus or minus so let's go ahead and um use our test point which was what one three again so if we plug that in here we want to get three when we plug in a positive one so this is e to the zero so again which one is it going to be it's going to be the positive so the solution again the solution whichever form you wanted to use is going to be the positive and this is what i was saying that you you have to put plus or minus and show you used a test point you can't just forget about the plus and minus and automatically get the positive answer you have to show them why you chose the positive not that you forgot the plus and the minus so be very careful with that okay so now it says be sure to state the domain of your solution equation um you have to be careful when you're looking at your solution equation what do you notice here x cannot equal zero so again another thing that might cost you a point not that big of a deal but those are the the nuances of this unit is the log properties simplifying i mean you're used to condensing to this but are you used to splitting this up um those are the things that might occur when you are when you see some problems in this unit not a major deal um but that's the hardest part okay and we are done with those two topics