Initial Value Problems Example 1 We’re going to move on and go ahead and do some initial value problems. They usually look like and there was basic form. The problem will give you two things dy or dx in an equation with this and at the same time basic initial value problems. You always get these two things. This is the derivative of the original function and this is a specific condition and they’ll ask you to solve the initial value problem. So when you do that, this notation here indicates that this is the first derivative of the original function. So whether that matters what you want to do is go ahead and integrate this function. So you’ll say the integral of 2x + 1 dx and solve that which is simple enough. The integral of this is x2 2/2 is 1 + x +c. So we have now the integral of the function and what we want to do is we put c here to account our constant. You have given this a condition so that we can solve for c and come up with a final equation so the way that we’re going to do that is use this condition here and what does that means is when you plug in zero for x the answer you get out is 3. So basically they’re giving you a point to satisfy this function and only you have to do is plug it in. So you’re going to say 3= and then plug in zero for x, 02 + 0 + c. So that’s why you do the plug in and then of course the answer here 3= 02 goes away, 0 goes away and you’ll have to say is C. so C equals 3 then all you have to do is write your final answer which is x2 + x which I’m getting from here before we plug in the zeros and then plus c which we solved for 3. So, if the problem asked you solve the initial value problem, this is the way you’re going to do it and this is your final answer in this case. We’ll go ahead and do some more examples.