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Verifying solutions to differential equations | AP Calculus AB | Khan Academy


– [Instructor] So let’s write
down a differential equation, the derivative of y with respect to x is equal to four y over x. And what we’ll see in
this video is the solution to a differential equation isn’t a value or a set of values. It’s a function or a set of functions. But before we go about
actually trying to solve this or figure out all of the solutions, let’s test whether certain equations, certain functions, are solutions to this
differential equation. So for example, if I have y is equal to four x, is this a solution to this
differential equation? Pause the video and see
if you can figure it out. Well to see if this is a solution, what we have to do is figure
out the derivative of y with respect to x and
see is that truly equal to four times y over x. And I’m gonna try to express
everything in terms of x to see if I really have an equality there. So first let’s figure
out the derivative of y with respect to x. Well that’s just going
to be equal to four. We’ve seen that many times before. And so what we need to test is, is four, the derivative
of y with respect to x, equal to four times, I could write y, but instead of y let’s write four x. I’m gonna put everything in terms of x. So y is equal to four x, so instead of four y I could
write four times four x, all of that over x. Is this true? Well that x cancels with
that and I’m gonna get four is equal to 16,
which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let’s look at another equation. What about y is equal to
x to the fourth power? Pause this video and see
if this is a solution to our original differential equation. Well we’re going to do the same thing. What’s the derivative
of y with respect to x? This is equal to, just
using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that’s the derivative
of y with respect to x, equal to four times y, instead of writing a y I’m gonna
write it all in terms of x, so is that equal to four
times x to the fourth, because x to the fourth
is the same thing as y, divided by x? And so let’s see, x to
the fourth divided by x, that is going to be x to the third. And so you will indeed
get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It’s not necessarily the only solution, but it is a solution to
that differential equation. Let’s look at another
differential equation. Let’s say that I had, and I’m gonna write it
with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let’s say I have f of x is equal to two x. Is this a solution to this
differential equation? Pause the video again and
see if you can figure it out. Well to figure that out, you have to say well what is f prime of x? f prime of x is just
going to be equal to two. And then test the equality. Is two, is f prime of x, equal to f of x, which is two x, minus x, minus x? And so let’s see we are going
to get two is equal to x. So you might be tempted to say oh hey I just solved for x or
something like that. But this would tell you
that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I’ll just put an x there, or I’ll put a incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a
function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let’s look at another one. Let’s say that we have f of
x is equal to x plus one. Pause the video and
see, is this a solution to our differential equation? Well same drill. f prime of x is going to be equal to one. And so we have to see is f prime of x, which is equal to one,
is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let’s do a few more of these. Let me scroll down little bit
so I have a little bit more, a little bit more space, but make sure we see our
original differential equation. Let’s test whether, I’m
gonna do it in a red color, let’s test whether f of x equals e to the x plus x plus one is a solution to this
differential equation. Pause the video again and
see if you can figure it out. All right well let’s figure
out the derivative here. f prime of x is going to be equal to, derivative of e to the x with
respect to x is e to the x, which I always find amazing. And so and then plus one
and the derivative of this with respect to x is just zero. And then let’s substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x
plus one, minus x, minus x? And if that x cancels out with that x, it is indeed, they are indeed equal. So this is also a solution. So this, this is a solution. And we’re done.

Bernard Jenkins

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