Uploaded by TheIntegralCALC on 26.02.2011

Transcript:

Hi everyone. Welcome back to integralCALC.com. Today we’re going to be doing another second-order

differential equations problem and in this one, they’ve asked us to solve this initial

value problem here. They’ve given us a function and two initial conditions and asked us to

solve the initial value problem.

So we’re going to need a couple of formulas first. We will need the product rule later

on. This is the second-order differential equations formula that we will need and it’s

a two-part formula that also involves this p plus or minus qi and then we will need the

quadratic formula in order to solve for r, roots, which you will see in a second.

So the first thing we want to do with this problem is convert our original function here

that involves the variable y to an equation that involves the variable r. So we’re going

to be replacing Ys with Rs and I think the easiest way to do it, it’s a quick little

trick that I use. I count the hash marks on each y variable. So let’s pull up what our

equation is going to look like.

So we’re going to replace each of the y variables with r but we have to decide what

our exponent is going to be on our y variables down here. So I count the hash marks. So here

we have y with two hash marks which means we’re going to have r squared and you can

see we replaced it with r squared here.

Here we have y with one hash mark so we’re going to replace that with r to the 1 or just

r and you can see we’ve replaced that there. Here for y, we have y with zero hash marks

so we’re going to replace that with r to the zero but r to the zero is just 1. So we

multiply 20 times 1. We just get 20. That kind of drops away but that’s how we replace

Ys with Rs and the reason that we’re going to do that is because we’re going to now

use this equation to solve for r and that will give us the roots or solutions to our

equation and we will be able to plug those into this formula here.

So we will normally try to factor this equation. In this case, we can’t easily factor it

so we’re going to need to use the quadratic formula here to solve for r and remember that

with the quadratic formula, a is the coefficient on your squared term so in our case 1; b is

the coefficient on r here so in our case that’s 4 and c is the third term here that doesn’t

have the variable attached to it. So in our case, that’s 20.

So we will be plugging in 1 for a, 4 for b and 20 for c into our quadratic formula here

and this is what we get and when we simplify a couple of steps, we will end up with negative

4 plus or minus the square root of negative 64 all over 2.

Now the first thing that we need to notice is that we have a negative number now inside

of our square root sign which means we’re going to end up with an imaginary solution

and we represent an imaginary solution with this variable i and remember that i is equal

to the square root of negative 1 when we’re talking about imaginary numbers or imaginary

solutions.

So we will convert the solution to the square root of 64 times the square root of negative

1. This here is the same as just the square root of negative 64 and the reason we do that

is because obviously when we take the square root of 64, we will get 8 and remember we

said that the square root of negative 1 was equal to i. So this whole thing here just

becomes i in our solution.

If we divide through by 2, r will end up equal to negative 2 plus or minus 4i. Now notice

that this solution is in the exact same form as this p plus or minus qi which was the second

part of this two-part formula here for second-order differential equations.

So the information we need to pull out of this solution is p and q. So you can see that

since they’re in the same form, negative 2 will be equal to p up here and 4 will be

equal to q. So we will now plug in negative 2 for p right here and 4 for q both here and

here and here’s what that looks like.

So now that we’ve solved for r and we’ve identified p and q and plugged those into

our formula for y of x, we found our y of x equation which is super important. So we’ve

gotten that far. The rest of the way, we’re going to be doing a couple of things.

First we’re going to take the derivative of y of x to get y prime of x and once we

do that, we will plug in our two initial conditions here, y of zero equals 9. This one will be

plugged into our original native y of x equation and we will be plugging in y prime of zero

equals 10 to y prime of x or the derivative of this equation. So we will match those up

and we will get two equations back that both involve c sub 1 and c sub 2 and with those

two equations, we will use a method of substitution or simultaneous equations to solve for c sub

1 and c sub 2; and whatever we get, we will plug back into this y of x equation here for

c sub 1 and c sub 2 and whatever we get will be our final answer.

So let’s go ahead and start off. So I guess the way I have it here, we will plug in the

initial condition first and then we will take the derivative and plug in its corresponding

initial condition. So in this case, we have y of x so we’re going to take this initial

condition here. We will be plugging in zero for x and we will set it equal to 9.

So when we plug in zero for x and we set it equal to 9, we notice a couple of things.

First, this first term here simplifies to e to the zero. Anything raised to the zero

power is 1. So this first term will reduce to 1 which we’ve written down here.

As for everything inside the parentheses, c sub 1 and c sub 2 will remain and to evaluate

cosine of zero and sine of zero, you can just use your calculator. The technically correct

way to do it would be to pull up your unit circle and look at that, cosine where the

angle is equal to zero on the unit circle. Remember that when we’re using the unit

circle, we’re looking at the angle zero. Cosine is going to be the x-coordinate of

that location and sine is going to be the y-coordinate at that location. So at the angle

zero, cosine, the x-coordinate is going to be equal to 1.

So in place of cosine of zero, we put a 1. Sine at the angle zero is going to be the

y-coordinate of that angle and in this case, the y coordinate is zero. So sine of zero

gets replaced with zero and simplifying that, you can see this whole term will go away.

This 1 outside is redundant. We’re just left with c sub 1 times 1 equals 9 and therefore

c sub 1 is equal to 9.

So we’ve already solved for c sub 1. We need to now take the derivative of y of x

to get y prime of x and use the second initial condition to solve for c sub 2.

So taking the derivative, it’s a little bit long. We’re going to need to use the

product rule which hopefully by now you’re familiar with but we will treat e to the negative

2x as our first function f of x here in product rule and everything here inside the parentheses

as our second function g of x.

When we do that, here’s what we’re going to get for the derivative. Remember that we

will take the derivative of f of x so that’s the derivative of e to the negative 2x which

is negative 2e to the negative 2x and then we leave g of x completely alone. So everything

inside the parentheses, leave that untouched.

Then we move on and we add f of x untouched. So e to the negative 2x stays as it is right

here and then we take the derivative of g of x, g prime of x. So we’re taking the

derivative of everything here inside the parentheses.

So, when we take the derivative of this part right here, remember that c sub 1 and c sub

2 are going to act like coefficients on these terms. So they’re going to stay right where

they are. They’re not like variables. They’re just like a coefficient in front of this cosine

of 4x and sine of 4x.

So we’re really looking at the derivative here of cosine of 4x and for that, we will

need to apply chain rule because we have 4x inside of cosine. So looking at the derivative

of cosine, the derivative of cosine is negative sine. So you can see we’ve brought this

negative sign out in front and then we have sine here. So, negative sine leaving the 4x

completely alone. This is chain rule, right? So we leave that 4x but then according to

chain rule, we have to multiply by the derivative of the inside part which is 4x. The derivative

of the inside 4x is 4 so we multiply by 4 and that’s where we get the 4 right there.

Again we will need to apply chain rule to the second term here. So treat sine as the

outside function and 4x as the inside function. So taking the derivative of the outside sine,

the derivative of sine is cosine. So you can see we have that and we leave the 4x completely

alone because we take the derivative of the outside. Leave the inside alone.

But then we have to multiply by the derivative of the inside which the inside is 4x so the

derivative is 4. So you can see we’ve multiplied by 4 here and we leave c sub 1 and c sub 2

alone as we said before.

So now this is going to be a bit of simplification. What we’re going to do – actually without

simplifying, we will just go straight to plugging in our initial condition. So y prime of zero

equals 10. That means that we’re going to plug zero in for x which you can see we’ve

done everywhere here and we’re going to set it equal to 10. So you can see we’ve

set it equal to 10 right here and now we’re going to go ahead and simplify. Remember that

anything raised to the zero power is 1. So e to the zero will be 1. We will end up with

negative 2 out in front.

Again for all of these sine and cosine, you will need the unit circle to evaluate or you

can just plug it into your calculator either way but cosine of zero is equal to 1. Sine

of zero is equal to zero. Again, this e to the zero is just 1 so that part drops away.

It would be redundant to write the 1 in front. Sine of zero is zero so we plug that in and

we just are left with this negative 4c sub 1 times zero and then cosine of zero again

is 1. So in place of cosine of 4 times zero, we plug in 1.

And now to simplify further, we will multiply this out. We will have negative 2 times c

sub 1 plus 4 times c sub 2. This term will drop away. This term will drop away here and

we will just be left with 4 times c sub 2. We set that equal to 10 and now, we will use

the fact that we solved for c sub 1 earlier. We found that it was 9. We will plug in 9

for c sub 1 and this is kind of that simultaneous equations part where we substitute in one

value so that we can solve just for c sub 2.

So when we multiply those together, we will get negative 18 plus 4c sub 2 equals 10. If

we add 18 to both sides, to try to get c sub 2 by itself, we will get 4c sub 2 equals 28

and then dividing both sides by 4 gives us c sub 2 equals 7.

So now that we’ve solved both for c sub 1 and c sub 2, we will plug them back into

our equation that we found earlier for y of x. Remember that we had previously solved

for both p and q. We said that p was equal to negative 2 which is how we got this coefficient

right here and that q was equal to 4 which is why we have 4x both inside of the cosine

function and the sine function.

Then for c sub 1 equals 9, c sub 1 was the coefficient here so we plug that in and then

c sub 2 was the coefficient on the sine term so we plug in 7 in place of c sub 2 and that’s

it.

This is our final answer. We leave it in this form. You can multiply or distribute the e

to the negative 2x if you want but it’s cleanest to leave it factored like this. So

I hope that helps you guys and I will see you in the next video. Bye.

differential equations problem and in this one, they’ve asked us to solve this initial

value problem here. They’ve given us a function and two initial conditions and asked us to

solve the initial value problem.

So we’re going to need a couple of formulas first. We will need the product rule later

on. This is the second-order differential equations formula that we will need and it’s

a two-part formula that also involves this p plus or minus qi and then we will need the

quadratic formula in order to solve for r, roots, which you will see in a second.

So the first thing we want to do with this problem is convert our original function here

that involves the variable y to an equation that involves the variable r. So we’re going

to be replacing Ys with Rs and I think the easiest way to do it, it’s a quick little

trick that I use. I count the hash marks on each y variable. So let’s pull up what our

equation is going to look like.

So we’re going to replace each of the y variables with r but we have to decide what

our exponent is going to be on our y variables down here. So I count the hash marks. So here

we have y with two hash marks which means we’re going to have r squared and you can

see we replaced it with r squared here.

Here we have y with one hash mark so we’re going to replace that with r to the 1 or just

r and you can see we’ve replaced that there. Here for y, we have y with zero hash marks

so we’re going to replace that with r to the zero but r to the zero is just 1. So we

multiply 20 times 1. We just get 20. That kind of drops away but that’s how we replace

Ys with Rs and the reason that we’re going to do that is because we’re going to now

use this equation to solve for r and that will give us the roots or solutions to our

equation and we will be able to plug those into this formula here.

So we will normally try to factor this equation. In this case, we can’t easily factor it

so we’re going to need to use the quadratic formula here to solve for r and remember that

with the quadratic formula, a is the coefficient on your squared term so in our case 1; b is

the coefficient on r here so in our case that’s 4 and c is the third term here that doesn’t

have the variable attached to it. So in our case, that’s 20.

So we will be plugging in 1 for a, 4 for b and 20 for c into our quadratic formula here

and this is what we get and when we simplify a couple of steps, we will end up with negative

4 plus or minus the square root of negative 64 all over 2.

Now the first thing that we need to notice is that we have a negative number now inside

of our square root sign which means we’re going to end up with an imaginary solution

and we represent an imaginary solution with this variable i and remember that i is equal

to the square root of negative 1 when we’re talking about imaginary numbers or imaginary

solutions.

So we will convert the solution to the square root of 64 times the square root of negative

1. This here is the same as just the square root of negative 64 and the reason we do that

is because obviously when we take the square root of 64, we will get 8 and remember we

said that the square root of negative 1 was equal to i. So this whole thing here just

becomes i in our solution.

If we divide through by 2, r will end up equal to negative 2 plus or minus 4i. Now notice

that this solution is in the exact same form as this p plus or minus qi which was the second

part of this two-part formula here for second-order differential equations.

So the information we need to pull out of this solution is p and q. So you can see that

since they’re in the same form, negative 2 will be equal to p up here and 4 will be

equal to q. So we will now plug in negative 2 for p right here and 4 for q both here and

here and here’s what that looks like.

So now that we’ve solved for r and we’ve identified p and q and plugged those into

our formula for y of x, we found our y of x equation which is super important. So we’ve

gotten that far. The rest of the way, we’re going to be doing a couple of things.

First we’re going to take the derivative of y of x to get y prime of x and once we

do that, we will plug in our two initial conditions here, y of zero equals 9. This one will be

plugged into our original native y of x equation and we will be plugging in y prime of zero

equals 10 to y prime of x or the derivative of this equation. So we will match those up

and we will get two equations back that both involve c sub 1 and c sub 2 and with those

two equations, we will use a method of substitution or simultaneous equations to solve for c sub

1 and c sub 2; and whatever we get, we will plug back into this y of x equation here for

c sub 1 and c sub 2 and whatever we get will be our final answer.

So let’s go ahead and start off. So I guess the way I have it here, we will plug in the

initial condition first and then we will take the derivative and plug in its corresponding

initial condition. So in this case, we have y of x so we’re going to take this initial

condition here. We will be plugging in zero for x and we will set it equal to 9.

So when we plug in zero for x and we set it equal to 9, we notice a couple of things.

First, this first term here simplifies to e to the zero. Anything raised to the zero

power is 1. So this first term will reduce to 1 which we’ve written down here.

As for everything inside the parentheses, c sub 1 and c sub 2 will remain and to evaluate

cosine of zero and sine of zero, you can just use your calculator. The technically correct

way to do it would be to pull up your unit circle and look at that, cosine where the

angle is equal to zero on the unit circle. Remember that when we’re using the unit

circle, we’re looking at the angle zero. Cosine is going to be the x-coordinate of

that location and sine is going to be the y-coordinate at that location. So at the angle

zero, cosine, the x-coordinate is going to be equal to 1.

So in place of cosine of zero, we put a 1. Sine at the angle zero is going to be the

y-coordinate of that angle and in this case, the y coordinate is zero. So sine of zero

gets replaced with zero and simplifying that, you can see this whole term will go away.

This 1 outside is redundant. We’re just left with c sub 1 times 1 equals 9 and therefore

c sub 1 is equal to 9.

So we’ve already solved for c sub 1. We need to now take the derivative of y of x

to get y prime of x and use the second initial condition to solve for c sub 2.

So taking the derivative, it’s a little bit long. We’re going to need to use the

product rule which hopefully by now you’re familiar with but we will treat e to the negative

2x as our first function f of x here in product rule and everything here inside the parentheses

as our second function g of x.

When we do that, here’s what we’re going to get for the derivative. Remember that we

will take the derivative of f of x so that’s the derivative of e to the negative 2x which

is negative 2e to the negative 2x and then we leave g of x completely alone. So everything

inside the parentheses, leave that untouched.

Then we move on and we add f of x untouched. So e to the negative 2x stays as it is right

here and then we take the derivative of g of x, g prime of x. So we’re taking the

derivative of everything here inside the parentheses.

So, when we take the derivative of this part right here, remember that c sub 1 and c sub

2 are going to act like coefficients on these terms. So they’re going to stay right where

they are. They’re not like variables. They’re just like a coefficient in front of this cosine

of 4x and sine of 4x.

So we’re really looking at the derivative here of cosine of 4x and for that, we will

need to apply chain rule because we have 4x inside of cosine. So looking at the derivative

of cosine, the derivative of cosine is negative sine. So you can see we’ve brought this

negative sign out in front and then we have sine here. So, negative sine leaving the 4x

completely alone. This is chain rule, right? So we leave that 4x but then according to

chain rule, we have to multiply by the derivative of the inside part which is 4x. The derivative

of the inside 4x is 4 so we multiply by 4 and that’s where we get the 4 right there.

Again we will need to apply chain rule to the second term here. So treat sine as the

outside function and 4x as the inside function. So taking the derivative of the outside sine,

the derivative of sine is cosine. So you can see we have that and we leave the 4x completely

alone because we take the derivative of the outside. Leave the inside alone.

But then we have to multiply by the derivative of the inside which the inside is 4x so the

derivative is 4. So you can see we’ve multiplied by 4 here and we leave c sub 1 and c sub 2

alone as we said before.

So now this is going to be a bit of simplification. What we’re going to do – actually without

simplifying, we will just go straight to plugging in our initial condition. So y prime of zero

equals 10. That means that we’re going to plug zero in for x which you can see we’ve

done everywhere here and we’re going to set it equal to 10. So you can see we’ve

set it equal to 10 right here and now we’re going to go ahead and simplify. Remember that

anything raised to the zero power is 1. So e to the zero will be 1. We will end up with

negative 2 out in front.

Again for all of these sine and cosine, you will need the unit circle to evaluate or you

can just plug it into your calculator either way but cosine of zero is equal to 1. Sine

of zero is equal to zero. Again, this e to the zero is just 1 so that part drops away.

It would be redundant to write the 1 in front. Sine of zero is zero so we plug that in and

we just are left with this negative 4c sub 1 times zero and then cosine of zero again

is 1. So in place of cosine of 4 times zero, we plug in 1.

And now to simplify further, we will multiply this out. We will have negative 2 times c

sub 1 plus 4 times c sub 2. This term will drop away. This term will drop away here and

we will just be left with 4 times c sub 2. We set that equal to 10 and now, we will use

the fact that we solved for c sub 1 earlier. We found that it was 9. We will plug in 9

for c sub 1 and this is kind of that simultaneous equations part where we substitute in one

value so that we can solve just for c sub 2.

So when we multiply those together, we will get negative 18 plus 4c sub 2 equals 10. If

we add 18 to both sides, to try to get c sub 2 by itself, we will get 4c sub 2 equals 28

and then dividing both sides by 4 gives us c sub 2 equals 7.

So now that we’ve solved both for c sub 1 and c sub 2, we will plug them back into

our equation that we found earlier for y of x. Remember that we had previously solved

for both p and q. We said that p was equal to negative 2 which is how we got this coefficient

right here and that q was equal to 4 which is why we have 4x both inside of the cosine

function and the sine function.

Then for c sub 1 equals 9, c sub 1 was the coefficient here so we plug that in and then

c sub 2 was the coefficient on the sine term so we plug in 7 in place of c sub 2 and that’s

it.

This is our final answer. We leave it in this form. You can multiply or distribute the e

to the negative 2x if you want but it’s cleanest to leave it factored like this. So

I hope that helps you guys and I will see you in the next video. Bye.