Original subtitles

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Daring to be talking about how to solve a basic initial value problem and in this particular problem

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we've been given the function D-y over DX which means the derivative of Y with respect to x is equal

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to 2 x plus 1.

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We've also been given what's called an initial condition Y of 0 equals 3.

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And what this tells us is that when we plug 0 into our original function y of X not this function.

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But the integral of this function when we plug 0 in for x into the integral of this function the value

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that we're going to get back is three.

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So we have that piece of information and we have the derivative of our original function.

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What we need to do is find the integral of this function.

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That'll give us our original function.

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Then we're going to use this initial condition to plug in and solve for an unknown constant.

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So the way that we're going to do this is by separating our variables y and x.

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So we want to get the Ys on the left hand side the X on the right hand side the way that we're going

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to do that is by multiplying both sides of this function here by DX when we do that we'll get d wine

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is equal to 2 x plus 1 times d X..

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Now that we have our variables separated with Y's on the left and X is on the right we want to go ahead

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and integrate both sides.

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So we'll say the integral of d y is equal to the integral of 2 x plus 1 D X like this.

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Now when we take the integral of the left hand side.

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Don't think of this as just the integral of D-y.

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Think of it as the integral of one times D-y.

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Remember that when we take the integral of a constant the result is that constant multiplied by the

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variable that we're integrating with respect to in other words 1 times why or just why the integral

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of 1 is why when we're taking the integral with respect to y.

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So the result of this left hand side over here is just going to be y the integral of one is y.

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Over on the right hand side we want to take the integral term by term.

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The integral of two x remember that will add 1 to the exponent the exponent right now is 1 this is X

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to the first power.

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When we add one will get X squared.

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But then we need to divide this term here by the new exponent which is 2 2 divided by 2 our existing

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coefficient 2 divided by our new exponent 2 is just going to give us 1 which we don't need to write

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as a coefficient in front of this x squared term here.

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So we only end up with X squared.

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The integral of one is just going to be X so will get plus X but then remember that whenever we take

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an indefinite integral we always have to add c to account for the constant of integration.

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So we're going to add c to the right hand side.

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Now a lot of times people ask me why do we add c to the right hand side and not to the left hand side.

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Well really the reason that we do it is just to take a shortcut.

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But technically if we were being really technically correct we would also add c to the left hand side

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and we would do this.

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We would say plus CSUB one maybe we would call it.

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And on the right hand side we'd say plus CSUB two we call them like that.

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You'll see why in a second we just bother adding it to the right hand side.

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It's really just a shortcut because we're always going to end up with the same answer.

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But what we would do at this point is we would say why is equal to.

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We're going to go ahead and subtract.

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See someone from both sides.

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So what will be left with is y equals x squared plus X plus.

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Now we have this value here CSUB Sabtu minus see someone because we subtracted see someone from both

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sides.

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Well CSUB to minus see someone.

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We've got two constants here.

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Right now we don't know the value of each.

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But let's say that the value of CSUB 2 is equal to 5 right CSUB two is equal to five.

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And let's say see someone is equal to to see someone is equal to two.

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And here we have the value CSUB to minus see sub 1.

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So we have five minus two and the value is three.

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Well as it turns out we don't specifically need to know the value of CSUB two different from the value

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of seis of one.

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We just need to know the value of whatever constant is here.

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So what we can do is replace this whole thing with just a single value of c will say Y is equal to x

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squared plus X plus C.

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We've essentially replaced both CSUB 2 and see someone with a generic.

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See that represents a constant.

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When we solve for this constant what we'll end up with is three.

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In this example If these were the actual values we'd end up with three.

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We don't need to solve for 5 and 2 we don't need to know these values separately.

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We just need to know this value in general.

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And having just a single see here is going to lead us to this value.

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So it's plenty adequate.

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And in fact we have to do it this way because we don't have enough equations to solve for CSUB to see

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someone separately we wouldn't we wouldn't be able to go about finding separate values for these two

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constants with this one initial condition.

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We're going to be able to find a single value for the constant c so that's why we combine them and that's

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why we only have to add a single C to the right hand side and we ignore adding it to the left hand side

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because we're always just going to eventually end up in this spot where we just have A C on the right

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hand side.

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So you can skip these steps and go straight to here and only add the C on the right hand side.

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Now once we've gotten to this point we move on to our initial condition and remember that we said this

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is essentially.

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Remember the function y of x is equal to some function over here on the right in terms of x.

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So this value because this is why of X this value here 0 is what we're going to be plugging in for x

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remember that we always plug this in for x 3 is going to be our resulting y value.

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So if we plug in zero for x we get 3 for y.

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So essentially what we're going to do is plug zero in for x and 3 in for Y.

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And when we plug those in all will be left with is Seiza we'll be able to solve for that value of c..

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OK.

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So a plug three and for y and will get three is equal to.

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Now here we plug in zero for X so we're going to get zero squared.

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So zero squared plus zero here goes in for x and then we just have our C leftovers we get plus C.

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Obviously here we're going to get zero to go away and zero to go away.

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And as you can see we're just left with a value for C of C equals 3.

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And now we just plug this value for c back into our function here that we found for Y.

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And our final answer for this initial value problem is y equals x squared plus X plus 3.

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The value here that we got for C.