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what we're going to do in this video see if we can solve the differential equation the derivative of Y with respect to X is equal to x times y pause this video and see if you can find a general solution here so the first thing that my brain likes to do when I see a differential equation is to say hey is this separable and when I say separable can I get all the Y's and dys on one side and all the XS and DX's on the other side and you can indeed do that if we treat our differentials like if we could treat them like algebraic variables which is fair game when you're dealing with differential equations like this we could multiply both sides by DX so multiply both sides by DX and we could divide both sides by Y let me move this over a little bit so we have some space so we could also multiply both sides by 1 over Y 1 over Y and so what that does is these DX is canceled out and this Y and 1 over Y cancels out and we are left with let me write all the things in terms of y on the left-hand side in blue so we have 1 over Y dy is equal to and then I'll do all this stuff in orange we have as equal to we're just left with an X and a DX X DX and then we'll want to take the indefinite integral of both sides now what's the antiderivative of 1 over Y well if we want it in the most general form this would be the natural log of the absolute value of y and then this is going to be equal to the antiderivative of X is x squared over 2 and then we want to put a constant on either side I'll just put it on the right-hand side plus C this is ensures that we're dealing with the general solution now if we want to solve explicitly for Y we could raise e to both sides power another way to think about it is if this is equal to that then e to this power is going to be the same thing as e to that power now what does that do for us well what is e to the natural log of the absolute you've why well I'm raising e to the power that I would have to raise e to to get to the absolute value of y so the left-hand side here it simplifies to the absolute value of y and we get that is being equal to now we could use our exponent properties this over here is the same thing as e to the x squared over 2 times e to the C times e to the C I am just using our exponent properties here well e to the C we could just view that as some other type of constant this is just some constant C so we could rewrite this whole thing as C e e to the x squared x squared over 2 hopefully you see what I'm doing there I just used my exponent properties e to the sum of two things is equal to e to the first thing times e to the second thing and I just said well e to the power of some constant C we could just real able that is let's call that our blue C and so this simplifies to blue C times e to the x squared over 2 now we still have this absolute value sign here so this essentially means that Y Y could be equal to we could write it this way Y could be equal to plus or minus c c/e e to the x squared over 2 x squared over 2 but once again we don't know what this constant is I didn't say that it was positive or negative so when you say a plus or minus of a constant here you could really just subsume all of this I'll just call this with red C so we could say that Y is equal to I'll just rewrite it over again for for fun y is equal to red C not the red sea but a red sea if times e to the x squared over 2 this right over here is the general solution to the original separable differential equation