in this video we're going to focus on solving differential equations by means of separating variables so let's start with this example problem dydx is equal to x^2 / y^2 now in this problem what we need to do is separate the variables on one side we only want to have y variables and on the other side of the equation we only want to have X variables so since we have a fraction separated by an equal sign Let's cross multiply Dy * y^2 is simply y^2 Dy and on the other side it's going to be x^2 DX now that we've separated the variables we can integrate both sides of the function the anti-derivative of y^2 is YB / 3 the anti-derivative of x^2 is XB / 3 and on one side of the equation you need to add a constant C so now let's get y by itself so let's multiply both sides of the equation by three so it's going to be y Cub is equal to xq + 3 C now 3 C is a constant so we can just write C for the total sum of all constants so you can rewrite this as y y Cub is equal to X Cub plus C and so this is the answer now if you want to you can take the cube root of both sides and simplify it further so Y is equal to the cube root of x Cub + C so this is the general solution now let's say if we want to find a particular solution if we're given an initial condition let's say y of 1 is equal to 2 if you're given a point an X and Y value you need to solve for c y is equal to 2 and X is equal to 1 so in order to get C by itself we need to raise both sides to the thir power 2 the 3 is equal to 8 the cube of a cube root cancels and it's just going to be 1 plus c so we could see that 7 is equal to C so the particular solution is y equal to the cube root of x Cub + 7 so this is the general solution and this is the particular solution given the initial condition let's work on another example let's say that y Prime is equal to x * y go ahead and pause the video and see if you can get this answer and let's say that the initial condition is y0 is equal to 5 you can also view this as the point 0 comma 5 x is z and y is 5 so let's begin y Prime is the same as dy over DX so we need to separate X and Y and let's begin by multiplying both sides by DX so therefore Dy is equal to XY DX now let's divide both sides by y so on the left what we have now is 1 y Dy and this is equal to X DX so now let's find the anti-derivative of both sides the anti-derivative of 1 y is the natural log of Y or I'm just going to write it as Ln Y and the anti-derivative of X is x^2 / 2 + C so now what do you think we should do at this point at this point what we can do is Place both sides of the function on the exponents of e if Ln y isal x^2 2 + C then then e to the Ln Y is equal to e 12 x^2 + C since the bases are the same then the exponents are equal to each other now e to the Ln y notice that natural log always have a base e these will cancel and so you just get y on the left side and that's equal to e raised to 12 x^2 + C now how can we separate X2 and C now it's important to understand that x^2 * X cube is equal to x 2 + 3 which is X 5th power whenever you multiply by Common bases you need to add exponents so therefore the reverse is true let's say if we have X raised to 4 + 7 we could separate 4 and 7 by rewriting as x 4 time x 7th the same way as we did it here so therefore we could separate the 12x^2 from C by means of multiplication so this is e 12x^2 time e to the C now e rais to the C is a constant so we could simply replace that with C therefore the general solution is y is equal to C * e raised to 12 x^2 now the last thing that we need to do is find a particular solution so we said that y of 0 is equal to 5 so let's find the value of the constant C so Y is 5 and x is 0 0 2 is 0 * a half that's Z so what we have is e raised to the 0 power anything raised to 0 power is equal to 1 so therefore we could see that c is equal to five so the particular solution is 5 e raised to the 12 x^2 and so that's the answer to this problem and here is the general solution here's another problem for you let's say that Dy / DX is equal to y^2 + 1 and let's say that you're given an initial condition F of one is equal to Z or rather y of one is equal to zero so go ahead and find the general solution and the particular solution to this differential equation so feel free to pause the video and work on this example now let's begin let's start by multiplying both sides sides by DX so what we now have is dy is equal to y^2 + 1 * DX so we need to move the Y variable to the other side and the best way to do that to separate it from DX is to divide both sides by y^2 + 1 let's rewrite it as one over y^2 + 1 * Dy and that's equal to DX so now let's find the anti-derivative of both sides so what is the anti-derivative of 1 y^2 + 1 well it helps to know that the derivative of the inverse tangent of X is equal to 1 over 1 + x^2 so this is going to be the anti-derivative I mean the anti-derivative of 1 y^2 + 1 is inverse tangent y so that's what we're going to have on the left side on the right side the anti-derivative of DX is simply 1 X plus C or just X plus c now what would you do at this point what we need to do is take the tangent of both sides of the equation that's how we can separate inverse tangent from y tangent and inverse tangent these two cancel so on the left side this is going to be Y is equal to tangent x + C so this is the general equation now we were given an initial point Y of 1 is equal to Z so let's replace y with zero and x with one and let's solve for C tangent of what angle is equal to one I mean zero not one tangent of 0 is equal to Z tangent of Pi is equal to Z as well but let's use this one so therefore 1 + C has to be equal to zero which means that c is equal to 1 so one of the particular Solutions is y is equal to tangent X - 1 now granted I guess you you can set 1 plus Cal to Pi you could do that for 2 pi 3 pi and it's just going to keep on going but I just chose one of those answers and here's one particular solution