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Incorporate equation resolution

in this video we're going to talk about how to solve first-order linear differential equations now the first thing you need to do is you need to write it in standard form and that is y prime plus P of x times y which is equal to Q of X now once you have it in standard form you need to identify the functions P of X and Q of X now the next thing you need to do is determine the integrating factor which is e raised to the integral of px DX now once you have the integrating factor you can now write the general solution to the differential equation so it's going to be 1 divided by the integrating factor times the integral of I of x times Q of X DX and then plus some constant of integration now once you integrate this portion to get the integrating factor and don't add the constant of integration you should only add it at the very end in this step so keep that in mind now let's work on some problems so let's say if we have this differential equation y prime plus 2y is equal to 2 e to the X so notice that this equation is already in standard form so if we write Y prime plus px times y which is equal to QX we can see that PX is equal to the stuff in front of the Y which is 2 and QX is just 2 e to the X so that's the second step to identify P of X and Q of X now let's determine the integrating factor IX so that's e raised to the integral P of X DX so that's going to be e times the integral of 2 DX the antiderivative of 2 DX is to X and so this is I of X so now we can write the general equation the general solution to the differential equation using this formula so it's going to be 1 over e to the 2x times the integral of e to the 2x and then Q is 2 e 3 X DX and n plus C now let's simplify the expression that we have before we integrate it so e to the 2x times e to the X what is that equal to so we need to add 2x plus 1 X which is to reiax so right now we have this expression we can move the 2 to the front so this is going to be 2 times the integral of e to the 3x 2 DX plus the cost of integration now the antiderivative of e to the 3x is e to the 3x divided by 3 so now let's distribute this to both terms so we're gonna have two over 3 times e to the 3x divided by e to the two x and then plus if we multiply these two C divided by e to the 2x...