Uploaded by TheIntegralCALC on 21.02.2011

Transcript:

Hi everyone! Welcome back to integralcalc.com. Today we’re going to be doing a second-order

differential equations problem but we’re going to be given initial values in this problem

so it’s going to be a combination of second order differential equations and initial values.

The equation that we’re given is two y double prime minus eleven y prime plus twelve y equals

zero and we’ve been given the initial conditions y of zero equals five and y prime of zero

equals fifteen. So, keep in mind that y double prime here represents the second derivative

of the original equation y and y prime represents the first derivative of the original equation

y. The formula that we’re going to be using

for this problem as with all second-order differential equations problem is this formula

here for y of x. So, just to give you kind of a preview of what we’re going to be doing,

we are going to be solving first for r sub one and r sub two here in the exponents. Once

we do that, we’re going to come up with an equation for y of x that still has c sub

one and c sub two in it. We will take the derivative of y of x to get y prime of x and

then we’ll use our initial conditions for y of zero and y prime of zero to plug in those

values, solve for c sub and c sub two, and, once we found those values we’ll plug them

in and we’ll come up with a final formula for y of x and that will be our final answer.

So, like I said, the first thing we’re going to do is solve for r sub one and r sub two.

And, the way that we do that is we convert our original equation here to an equation

with r’s in it instead of y’s and the way that we do that, simply, if we have y

here we treat it us one, if we have y prime we treat it as r, and if we have y double

prime we treat it as r squared. The easiest thing for me to remember is to just count

the hash marks on the y’s. If you have zero hash marks here, right, we have y with zero

hash marks, that’s r to the zero which is one, anything raised to zero power is one,

so we would multiply twelve by… twelve times one and we would just get twelve. Here, y

prime, we have one hash mark that means it’s r to the one which is just r and we plug in

r. We’ve got two hash marks here for y double prime, that’s r squared and we end up here

with r squared. So it’s a stupid trick, it’s not mathematical but it’s… makes

it easy for me to remember what is one, what’s r, and what’s r squared when I plug those

into the equation here. So, once I do that, like I said, we’re going to solve for r

and in this case we can factor our equation to two r minus three times r minus four, if

you can’t factor you can try completing the square or using the quadratic formula,

but either way you want to solve for r and obviously we’re going to end up with two

factors here. So, what we do is we set each of these factors equal to zero. So we’ll

start with two r minus three equals zero, we add three to both sides and we get two

r equals three, and then we divide both sides by two and we get r equals three halves. Now

we’re going to do the same thing with r minus four equals zero, we’re taking this

factor here, sending it equal to zero, obviously we just add four to both sides and we get

r equals four. So those are our two solutions for r, as you can see here, three halves and

four, and what we’re going to do is plug those into our formula for r sub one and r

sub two. It doesn’t particularly matter which order you do. I usually take the…

the smaller factor and plug it in for r sub one. So, I have r sub one equal to three halves

and r sub two equal to four. So this is our… this is our equation here

for y of x. Now, what we want to do is take the derivative of y of x and the way that

we do that… we treat c one and c two as a coefficient on this e to the x term so they’re

going to stay right where they are as coefficients. Remember that the derivative of e to some

constant times x, right, this three halves is the constant, it’s the coefficient on

the x term right here, taking the derivative of that kind of a term it means that we just

multiply this three halves out in front of the… of the term leaving the exponent three

halves x completely intact. So, all we do is bring three halves out in front, the c

sub one stays right where it is, and the exponent remains unchanged. Same thing here with second

term, we bring the four out in front and the c sub two stays right where it is as this,

the exponent four x. So that is the derivative of y of x, so its y prime of x.

Now we’re going to need both of these equations. We’re going to use our initial conditions

here. And, for this first initial condition, y of zero equals five, we will plug in zero

to our equation for y of x, we’ll plugging in four x to y of x and we’ll set it equal

to five and then for our second initial condition y prime of zero equals fifteen we will plug

zero in for x to y prime, the derivative, and set that equal to fifteen. We’ll come

out with two equations that both include c sub one and c sub two and then we’ll use

simultaneous equations to solve for c sub one and c sub two.

So let’s… let’s bring these two equations in here, these are the ones that we had just

found, y of x and y prime of x, and now let’s start working with our initial conditions.

So you can see here, we’re working with y of x, the original y of x equation that

we found here, and we’re going to plug in zero for x. So you can see we have y of zero,

we’re plugging in zero for x, and we’re setting it equal to five as our initial condition

indicates that we should do so we just plug in zero, we set it equal to five. Remember

that anything raised to the zero power is one, which means that we’ll end up with

c sub one times one plus c sub two times one is equal to five. Obviously, when we simplify

that we get c sub one plus c sub two equals five. So that’s our first equation, we’ll

keep that in the back of our mind, we move on to the second initial condition and then

we’ll bring this back later to solve for c sub one and c sub two using simultaneous

equations. So now, taking our second initial condition, we will plug in zero for x to y

prime and set that equal to fifteen. So you can see y prime of zero, we’ll plugging

in zero for x here and here and setting it equal to fifteen, just as we did with y of

zero equals five. So, again, anything raised to the zero power is one, so this e to the

three halves times zero just becomes one and we’re left with three halves c sub one times

one and then again four plus, or sorry, plus four times c sub two times one. Obviously,

when we simplify, these ones go away and we’re left with three halves c sub one plus four

c sub two equals fifteen. So these two here, c sub one plus c sub two equals five and this

one we just found, are the simultaneous equations that we’ll use to solve for c sub one and

c sub two. Let’s go ahead and work with our second

equation. We’re going to multiply all of the terms by two to get… just so that I

can get rid of the fraction. We’re going to end up with three times c sub one, this

will become eight c sub two, and then two times fifteen will be thirty. And then, we’ll

bring back our first equation that we got from y of zero equals five, c sub one plus

c sub two equals five, we’ll subtract c sub two from both sides to solve for c sub

one and then plug in this five minus c sub two into our other equation here for c sub

one. You can see we take this part here and plugged it in for this c sub one so that we

can have just c sub two in this equation and solve for it like a variable. So, distributing

the three here, three times five gives us fifteen minus three times c sub two here,

obviously, the rest of the equations stays the same. And then, when we simplify this,

negative three c sub two plus eight c sub two is a positive five c sub two and when

we subtract fifteen from both sides we get that five c sub two is equal to fifteen. Dividing

both sides by five gives us c sub two equal to three.

So now that we’ve solved for c sub two, we can plug it back in to our first equation

here. We’re going to plug in for c sub two and we’ll get c sub one plus three equals

five. When we subtract three from both sides, we’ll get c sub one equals two.

So that’s how you solve for both of those variables and remember, from before, we had

our equation already with r sub one and r sub two filled in, well, now we’re going

to take that same one and plug in c sub one and two to get our final answer. So, we had

plugged in three halves for r sub one, we found c sub one to be two so we plug it in

here. We plug in three for c sub two and four was what we came up with for r sub two and

so that carried over into our final answer here. So, this is the equation y of x that

ends up being our final answer where we’ve solved for both c sub one and two and r sub

one and two. I hope that it simply helped you guys and

I’ll see you in the next video. Bye!

differential equations problem but we’re going to be given initial values in this problem

so it’s going to be a combination of second order differential equations and initial values.

The equation that we’re given is two y double prime minus eleven y prime plus twelve y equals

zero and we’ve been given the initial conditions y of zero equals five and y prime of zero

equals fifteen. So, keep in mind that y double prime here represents the second derivative

of the original equation y and y prime represents the first derivative of the original equation

y. The formula that we’re going to be using

for this problem as with all second-order differential equations problem is this formula

here for y of x. So, just to give you kind of a preview of what we’re going to be doing,

we are going to be solving first for r sub one and r sub two here in the exponents. Once

we do that, we’re going to come up with an equation for y of x that still has c sub

one and c sub two in it. We will take the derivative of y of x to get y prime of x and

then we’ll use our initial conditions for y of zero and y prime of zero to plug in those

values, solve for c sub and c sub two, and, once we found those values we’ll plug them

in and we’ll come up with a final formula for y of x and that will be our final answer.

So, like I said, the first thing we’re going to do is solve for r sub one and r sub two.

And, the way that we do that is we convert our original equation here to an equation

with r’s in it instead of y’s and the way that we do that, simply, if we have y

here we treat it us one, if we have y prime we treat it as r, and if we have y double

prime we treat it as r squared. The easiest thing for me to remember is to just count

the hash marks on the y’s. If you have zero hash marks here, right, we have y with zero

hash marks, that’s r to the zero which is one, anything raised to zero power is one,

so we would multiply twelve by… twelve times one and we would just get twelve. Here, y

prime, we have one hash mark that means it’s r to the one which is just r and we plug in

r. We’ve got two hash marks here for y double prime, that’s r squared and we end up here

with r squared. So it’s a stupid trick, it’s not mathematical but it’s… makes

it easy for me to remember what is one, what’s r, and what’s r squared when I plug those

into the equation here. So, once I do that, like I said, we’re going to solve for r

and in this case we can factor our equation to two r minus three times r minus four, if

you can’t factor you can try completing the square or using the quadratic formula,

but either way you want to solve for r and obviously we’re going to end up with two

factors here. So, what we do is we set each of these factors equal to zero. So we’ll

start with two r minus three equals zero, we add three to both sides and we get two

r equals three, and then we divide both sides by two and we get r equals three halves. Now

we’re going to do the same thing with r minus four equals zero, we’re taking this

factor here, sending it equal to zero, obviously we just add four to both sides and we get

r equals four. So those are our two solutions for r, as you can see here, three halves and

four, and what we’re going to do is plug those into our formula for r sub one and r

sub two. It doesn’t particularly matter which order you do. I usually take the…

the smaller factor and plug it in for r sub one. So, I have r sub one equal to three halves

and r sub two equal to four. So this is our… this is our equation here

for y of x. Now, what we want to do is take the derivative of y of x and the way that

we do that… we treat c one and c two as a coefficient on this e to the x term so they’re

going to stay right where they are as coefficients. Remember that the derivative of e to some

constant times x, right, this three halves is the constant, it’s the coefficient on

the x term right here, taking the derivative of that kind of a term it means that we just

multiply this three halves out in front of the… of the term leaving the exponent three

halves x completely intact. So, all we do is bring three halves out in front, the c

sub one stays right where it is, and the exponent remains unchanged. Same thing here with second

term, we bring the four out in front and the c sub two stays right where it is as this,

the exponent four x. So that is the derivative of y of x, so its y prime of x.

Now we’re going to need both of these equations. We’re going to use our initial conditions

here. And, for this first initial condition, y of zero equals five, we will plug in zero

to our equation for y of x, we’ll plugging in four x to y of x and we’ll set it equal

to five and then for our second initial condition y prime of zero equals fifteen we will plug

zero in for x to y prime, the derivative, and set that equal to fifteen. We’ll come

out with two equations that both include c sub one and c sub two and then we’ll use

simultaneous equations to solve for c sub one and c sub two.

So let’s… let’s bring these two equations in here, these are the ones that we had just

found, y of x and y prime of x, and now let’s start working with our initial conditions.

So you can see here, we’re working with y of x, the original y of x equation that

we found here, and we’re going to plug in zero for x. So you can see we have y of zero,

we’re plugging in zero for x, and we’re setting it equal to five as our initial condition

indicates that we should do so we just plug in zero, we set it equal to five. Remember

that anything raised to the zero power is one, which means that we’ll end up with

c sub one times one plus c sub two times one is equal to five. Obviously, when we simplify

that we get c sub one plus c sub two equals five. So that’s our first equation, we’ll

keep that in the back of our mind, we move on to the second initial condition and then

we’ll bring this back later to solve for c sub one and c sub two using simultaneous

equations. So now, taking our second initial condition, we will plug in zero for x to y

prime and set that equal to fifteen. So you can see y prime of zero, we’ll plugging

in zero for x here and here and setting it equal to fifteen, just as we did with y of

zero equals five. So, again, anything raised to the zero power is one, so this e to the

three halves times zero just becomes one and we’re left with three halves c sub one times

one and then again four plus, or sorry, plus four times c sub two times one. Obviously,

when we simplify, these ones go away and we’re left with three halves c sub one plus four

c sub two equals fifteen. So these two here, c sub one plus c sub two equals five and this

one we just found, are the simultaneous equations that we’ll use to solve for c sub one and

c sub two. Let’s go ahead and work with our second

equation. We’re going to multiply all of the terms by two to get… just so that I

can get rid of the fraction. We’re going to end up with three times c sub one, this

will become eight c sub two, and then two times fifteen will be thirty. And then, we’ll

bring back our first equation that we got from y of zero equals five, c sub one plus

c sub two equals five, we’ll subtract c sub two from both sides to solve for c sub

one and then plug in this five minus c sub two into our other equation here for c sub

one. You can see we take this part here and plugged it in for this c sub one so that we

can have just c sub two in this equation and solve for it like a variable. So, distributing

the three here, three times five gives us fifteen minus three times c sub two here,

obviously, the rest of the equations stays the same. And then, when we simplify this,

negative three c sub two plus eight c sub two is a positive five c sub two and when

we subtract fifteen from both sides we get that five c sub two is equal to fifteen. Dividing

both sides by five gives us c sub two equal to three.

So now that we’ve solved for c sub two, we can plug it back in to our first equation

here. We’re going to plug in for c sub two and we’ll get c sub one plus three equals

five. When we subtract three from both sides, we’ll get c sub one equals two.

So that’s how you solve for both of those variables and remember, from before, we had

our equation already with r sub one and r sub two filled in, well, now we’re going

to take that same one and plug in c sub one and two to get our final answer. So, we had

plugged in three halves for r sub one, we found c sub one to be two so we plug it in

here. We plug in three for c sub two and four was what we came up with for r sub two and

so that carried over into our final answer here. So, this is the equation y of x that

ends up being our final answer where we’ve solved for both c sub one and two and r sub

one and two. I hope that it simply helped you guys and

I’ll see you in the next video. Bye!