 Let's take a look at the solution to question one from the final exam for our course math 1220 So the function y equals 5 minus c times e to negative x is a general solution for which of the following differential equations In which case then six options are given to us So what strategy could we take to try to determine if we have a solution for differential equation or not? Well, you have six differential equations, which we see right here Notice that they're all first order differential equation. They're all separable and most of them are linear differential equations We could just solve each and every one of them all six of them until we find this one That's not the suggestion. I would make but that is a that is a feasible strategy Instead the idea is much like if you had an algebraic equation like 2x plus 1 equals 5 is X equals 3 a solution to something like that right you could check it in you could plug it in there And you get 2 times 3 which is 6 plus 1 is 7 that doesn't quite work We could try x equals 2, you know We could we could plug those in there and that's what we can do with a differential equation as well And so notice that if I was just going to plug it into the first one, right? Let's compute the derivative of y with respect to x to give myself a little bit more space I'm going to do it over here So if we calculate the derivative y prime, this is going to equal 5 minus c e to the negative x prime derivative of a constant is 0 and then by the chain rule The derivative y prime is going to equal a negative negative c e to the negative x That is y prime equals positive c e to the negative x. So this is something to keep track of this is the left-hand side This is the left-hand side of this first equation right here on the other hand if I were to try to compute negative y plus 5 you're plugging y in here You're going to end up with negative times 5 minus c e to the negative x plus 5 Combining like terms the two fives combined together to actually give me a 10 negative 10 plus c e to the negative x and so this will be the right-hand side of that differential equation Do those match up with each other? The answer of course is no, right? We have c e to negative x We have a c e to negative x. We actually have an we have this negative 10. That's there We don't want that and so try to and so we try the first one We see it didn't match up right and so at the very least we're like hey The answer is not a and we could go through the other ones But we can also be a little bit more strategic with with this right notice that every option involves a y prime on the left Y prime y prime y prime so we don't have to compute y prime again We've already done that and now look at what's going on here since y prime is c e to the negative x What are some options we got here when we looked at this one y plus 5? You're gonna get that the 5 from y plus this way I've actually been combined to give us a 10 So you can't have a y plus 5 you need that the 5 actually disappears And so that also tells us that option f couldn't work either y plus 5 would give us a 10 So we're gonna have to have some type of subtraction right, but then the other thing to pay attention to is That we need to have a positive c e to the negative x Y starts off with a negative so we have to somehow turn that negative into a positive That is we have to be subtracting a y of some kind so we have to subtract from 5 the y and so we see that option d wouldn't work and Option e wouldn't work because we need to subtract the y from the 5 And so then we get that limbs it down just a choice b and c knows how quickly we're able to go through that If we tried b you take 5 minus 5 minus c e to the negative x Bob's your uncle you're gonna get 5 minus 5 they cancel and you're gonna get negative negative c's you get c e to negative x And so we see that the correct choice would be b right here One could also try c If you did c in that situation just just not that we need to do it But just so you're aware if we were trying c you'd end up with 1 over c e to the negative x which could simplify to be c e to the x within the exponent Is in rhyming in negative x right there So you know c didn't work as well and b gives you the correct answer once you find the correct So one only one of them is gonna work you can stop right there and go forward from that And so that's how one is expected to answer a question like number one just plug the equation Plug your solution your potential solution into the differential equations and see which one happens plug them all in and only one of Them's gonna work this one doesn't actually I mean again You could try solving each differential equation separately that'd be very very cost expensive And that's also assuming you'll be given a differential equation can solve So this one just wants to test whether you know to chat a check whether we have a solution to a differential equation or not