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Integrate initials deal

in this video I'm going to be doing some initial value problem before we actually get into the problem it's important that we go over differential equation so a differential equation is the following dy/dx equals a certain function f of headaches and now I can think it think of it as like normally be like to take the derivative of a function 250 F prime axes and this is the exact opposite of that so in this case we want to find the function that gives us the derivative of the following so now we're given the derivative and we need to find the original function so normally how you would solve these differential equations you would integrate both sides of the equation just like how if you want to add a number to an equation if that up to both sides you can integrate both sides respect to DX and you can find the original function for example if I were to solve this I would do the integral of dy DX their respective DX equals the integral of f of X is equal to DX and this would give us the solution y equals big f of X plus C and C is the sea is important because remember when I take the derivative of a constant it was 0 so if we take the derivative of this will definitely give this but this C can be anything so whenever we do a differential equation after we integrate both sides you always have to add plus C on the right-hand side so again differential equation is the exact opposite of taking the derivative and only we're given the derivative and we need to find the original function so initial value problems are all about finding that plus C after we get the general solution of the differential equation so again what we need is we need to find an equation that satisfies both of these Commission's or the first condition is that v yd x equals 10 minus x the second one is that why y of zero equals 1 so again if you remember to solve this differential equation we need to integrate both sides so let's do that so they're taking the integral of dy DX respect to DX and are making the integral of 10 minus x respect to DX a bit integral of dy DX is just Y and the integral of 10 minus x respect to DX is 10x minus x squared over 2 and remember we cannot react you cannot forget that plus C involved so if they include that as well and this is our general solution however what we need to find is we need to find a specific solution that sets with both of these conditions and we figured out we did this condition right here but we need to satisfy that condition and that says y is 0 equals 1 so we need to find what that...