Hi, everyone. Welcome back to integralcalc.com. Today, we’re going to be doing a linear
differentials problem. And in this one, we’ve been give the linear differential equation,
dy/dx plus 3y equals 2x times e to the negative 3x. So this here is our original problem and
the first thing that we need to confirm is that if it is in fact a linear differential
problem. And the way that we do that is by making sure that it is in the firm of this
first formula here. So in order to be a linear differential equation,
we need to have first of all dy/dx here on the left hand side and of course you can see
that we do in our problem. So that’s the first component. The second component is that
you need y here on the left hand side and it has to be y to the first power. Not y squared
or y cubed but just y. So we’ve got that and it can be multiplied by anything you want.
It can have an x variable in it but it needs to be either a constant or have an x variable
but obviously, you can’t have a y variable in that part because that would make this
term a y squared or y cubed. So that 3 there in our original problem is going to be p of
x which basically we consider to be the coefficient on the y term. And then over on the right-hand
side, you’ve got q of x. and q of x can be any function of x and it can be a constant.
It just can’t have a y variable involved in it. And our term on the right-hand does
in fact satisfy those requirements. We’ve got x variables involved here but we don’t
have y variables involved so we’re good. So you can see how these two things match
up with one another. So we can see by comparing our equation to the formula that we are in
fact dealing with a linear differential equation. so that’s the first step.
The second step is we have to calculate the integrating factoring which is given here
by this formula. So this is the notation that we use to denote integrating factor. And what
we have to do is find the integral of p of x and remember that we have p of x in our
original equations as 3. So we have to take the integral of p of x and then raise that
to the base e or take e to the power of whatever we’d get. So our integrating factor is going
to be e raised to the integral of p of x which we already know to be 3 so we’ve got 3
dx. The integral of 3 is 3x which means that our integrating factor is going to be equal
to e to the 3x. so we found our integrating factor. Once you find the integrating factor,
you need to go ahead and multiply it by both sides of your original linear differential
equation. so we can multiply it by every single term on both sides of our equation so what
we’ll get is e to the 3x times dy/dx plus e to the 3x times 3y equals 2x e to the -3x
times e to the 3x. so we just multiplied our integrating factor by every term in our equation.
Now that we’ve done that, we need to do a couple things. First of all, we need to
clean up the right-hand side of our equation because there’s some simplifying we can
do. But more importantly we need to recognize that the left-hand side of our equation is
the derivative of the product of two functions. Remember product rule from calc 1 where we
had f of x times g of x and we were asked to take the derivative of that and when we
did we found that the derivative was f prime of x times g of x plus f of x times g prime
of x. and that was what product rule was. That was how we took the derivative of two
functions that were multiplied together like this. The left-hand side of our equation follows
that same format. And what we can see is that we have one function which is y so we can
call that f of x. and the derivative of y is dy/dx. So that means that the derivative
of f of x is going to be dy/dx. So that’s one of them. And this kinds of take practice
to pick these things apart but we know that the derivative of e to the 3x is 3e to the
3x. So that means that g of x is e to the 3x and the derivative of g of x is 3e to the
3x so we can identify these things, take that left-hand side apart and realize that we can
rewrite the left-hand side as the multiplication of f of x and g of x so we know that f of
x is y and that g of x is e to the 3x. so we can say that the left-hand side is going
to be the derivative of e to the 3x times y. so we go ahead and write that. Then on
the right-hand side, since we have e to the negative 3x and e to the 3x, when the
base of your terms are the same, you can add their exponents together. So negative 3x plus
3x is of course zero. e raised to the zero power is 1. Anything raised to the zero power
is 1. So these cancel and we’re just left with 2x over here on the right hand side.
So now that we’ve done that, what we need to do is go ahead and integrate both sides
of the equation. and this is just linear differential equations process. These are the steps that
you’re always going to follow in terms of finding the integrating factor, then multiplying
through by the integrating factor and then writing the left-hand side of your equation
as the product of two functions, the derivative of the product of two functions. Now we’re
going to go ahead and integrate both sides. So we’re going to take the integral of d
of x of e to the 3x times y dx equals the integral of 2x dx. And when we take the integral
of a derivative because remember that this notation here means derivative; and then of
course, we’re taking the integral of that derivative, when you take the integral of
a derivative, that means the integral and the derivative are going to cancel with one
another so in the left-hand side here, we’re just going to be left with e to the 3x times
y. On the right-hand side, if we take the integral of 2x, we’ll just get x squared
and we want to go ahead and add the constant of integration to the right-hand side. So
once we’ve done that, the last thing we have to do in order to come up with a general
solution is solve for y. we want to get y on it’s own on the left-hand side. This
one’s pretty simple. We can just go ahead and multiply the entire equation both left
and right-hand side by e to the negative 3x because remember when we do that, when we
have e to the 3x times e to the negative 3x, we can add the exponents together. 3x plus
a negative 3x is a zero which means that this term over here is going to cancel and get
y on its own. So we end up with y on the left-hand side is equal to e to the negative 3x times
x squared plus c. And that’s our final answer. This is the general solution to our linear
differential equation. So that’s how we go about arriving at the general solution.
I hope that video helped you guys and I will see you in the next one. Bye!