Initial Value Problem Example 7
Uploaded by TheIntegralCALC on 19.09.2010
Transcript:
Hi everyone. Welcome back to integralcalc.com. Today we're going to be doing another initial
value problem and this one actually involves a trigonometric identity. So the problem we're
given is dy over dx equals y times secant squared of x and the initial condition is
y of zero equals five. So this is going to be a separable equation
because we have y and x on the right side as well as the left. So as with initial
value problem, the first thing that we're going to need to do is separate the variables
we're going to try to get the y's on the left side and the x's on the right side. So we're
going to start doing that by multiplying both sides with dx. So we'll have dy equals y times
secant squared of x dx and then we'll divide both sides by y so that we'll send all the
y's over here and the x's on the right side. So we'll end up with dy over y equals secant
squared of x dx. So now that we've separated the variables,
we will go ahead and integrate both sides and that will look like this. We just tap
on integrals to the front of both the left and the right side. And I'm going to go ahead
and convert the left side of this to be one over y times dy. It's a little bit easier
to see what we'll need to do that way but its actually the same thing right because
this is, we multiply this together we have dy on top here over y, so the same thing.
We just separating out the dy, equals the integral of secant squared of x dx.
So those are our integrals in their simple forms and now we're gonna go ahead and integrate.
the left side should be pretty obvious, whenever we have one over the variable, the integral
of that is natural log or ln of the variable so the left side ends up being ln of y. And
then the right side, this is a formula that you should memorize. The integral of secant
squared of x is actually just tangent of x and then we'll go ahead and add c to the right
side as we always do to account for the constant. So we've gone ahead and taken the integral
of both the left and right side and now what we need to do is try to get y on its own on
the left side and the way that we're going to do that, obviously, we're going to get
y by itself we need to get rid of the ln or the natural log, we're going to raise both
sides to base e. So ln of y becomes the exponent and tangent of x plus c is going to become
the exponent because we're going to raise both sides here to base e. So when we do that
and the reason we do that is because e to the natural log of something cancels out the
e and the natural log and we end up with only y on the left side. So you have y equal to
e to the tan x plus c. So that's the first part of our simplification
but of course we need to simplify the right side a little bit more and the way that we're
going to do that is by separating this c on the exponent. Whenever we have e to the a
plus b, whenever we have an exponent on the e, it's the addition or the sum of two different
terms as we do here, tan x and c, we can separate that to the e to the a times e to the b. So
here we can do the same thing. We’ll end up with y equals e to the tangent of x times
e to the c. That's just a formula that we need to know.
So now that we've done this and the reason that we do this is because e to the c is just
a constant and our c in here is acting as an area where we can consolidate all of our
constants. So because this is just a constant, it's the same thing as saying e to the tangent
of x times a constant so this is going to become the coefficient. It’s going to move
out here and we're going to change e to the c to simply c because c is accounting for
the constant in this function. So we'll end up with y equals c e to the tan x.
And we're actually done simplifying at this point and now that we've got our function
to this point, what we could go ahead and do is use the values in our initial condition,
plug them into this function and then solve for c, which is the end goal here. So we'll
go ahead and plug zero in for x and five in for y and solve for c. So we get five c e
to the tangent of zero. Tangent of zero, you can plug that in to your calculator if you
don't know it but it's just zero so we'll end up with five equals c e to the zero. Anything
raised to the zero power is just one so we'll end up with five equals c times one. And therefore
we have five equals to c. Now that we determined that c is five, we can go ahead and plug five
in for c and we'll get y equals five e to the tan x and that's our final answer.
So I'll see you guys next time. Thanks for watching.
value problem and this one actually involves a trigonometric identity. So the problem we're
given is dy over dx equals y times secant squared of x and the initial condition is
y of zero equals five. So this is going to be a separable equation
because we have y and x on the right side as well as the left. So as with initial
value problem, the first thing that we're going to need to do is separate the variables
we're going to try to get the y's on the left side and the x's on the right side. So we're
going to start doing that by multiplying both sides with dx. So we'll have dy equals y times
secant squared of x dx and then we'll divide both sides by y so that we'll send all the
y's over here and the x's on the right side. So we'll end up with dy over y equals secant
squared of x dx. So now that we've separated the variables,
we will go ahead and integrate both sides and that will look like this. We just tap
on integrals to the front of both the left and the right side. And I'm going to go ahead
and convert the left side of this to be one over y times dy. It's a little bit easier
to see what we'll need to do that way but its actually the same thing right because
this is, we multiply this together we have dy on top here over y, so the same thing.
We just separating out the dy, equals the integral of secant squared of x dx.
So those are our integrals in their simple forms and now we're gonna go ahead and integrate.
the left side should be pretty obvious, whenever we have one over the variable, the integral
of that is natural log or ln of the variable so the left side ends up being ln of y. And
then the right side, this is a formula that you should memorize. The integral of secant
squared of x is actually just tangent of x and then we'll go ahead and add c to the right
side as we always do to account for the constant. So we've gone ahead and taken the integral
of both the left and right side and now what we need to do is try to get y on its own on
the left side and the way that we're going to do that, obviously, we're going to get
y by itself we need to get rid of the ln or the natural log, we're going to raise both
sides to base e. So ln of y becomes the exponent and tangent of x plus c is going to become
the exponent because we're going to raise both sides here to base e. So when we do that
and the reason we do that is because e to the natural log of something cancels out the
e and the natural log and we end up with only y on the left side. So you have y equal to
e to the tan x plus c. So that's the first part of our simplification
but of course we need to simplify the right side a little bit more and the way that we're
going to do that is by separating this c on the exponent. Whenever we have e to the a
plus b, whenever we have an exponent on the e, it's the addition or the sum of two different
terms as we do here, tan x and c, we can separate that to the e to the a times e to the b. So
here we can do the same thing. We’ll end up with y equals e to the tangent of x times
e to the c. That's just a formula that we need to know.
So now that we've done this and the reason that we do this is because e to the c is just
a constant and our c in here is acting as an area where we can consolidate all of our
constants. So because this is just a constant, it's the same thing as saying e to the tangent
of x times a constant so this is going to become the coefficient. It’s going to move
out here and we're going to change e to the c to simply c because c is accounting for
the constant in this function. So we'll end up with y equals c e to the tan x.
And we're actually done simplifying at this point and now that we've got our function
to this point, what we could go ahead and do is use the values in our initial condition,
plug them into this function and then solve for c, which is the end goal here. So we'll
go ahead and plug zero in for x and five in for y and solve for c. So we get five c e
to the tangent of zero. Tangent of zero, you can plug that in to your calculator if you
don't know it but it's just zero so we'll end up with five equals c e to the zero. Anything
raised to the zero power is just one so we'll end up with five equals c times one. And therefore
we have five equals to c. Now that we determined that c is five, we can go ahead and plug five
in for c and we'll get y equals five e to the tan x and that's our final answer.
So I'll see you guys next time. Thanks for watching.