0:23

We know that e to the x satisfies corresponding homogeneous equation.

Say xy double prime- 1 + x

of y prime + y = 0, okay?

So corresponding homogeneous problem

is x y double prime- 1 + x y prime + y = 0.

This is second order differential equation and the variable coefficients, right?

The problem gives one solution y1 = e to the x,right?

This is the solution problem claims and you can confirm.

e to the x satisfies this homogeneous equation Very easily, right?

But because this given equation is of second order,

1:46

To find this general solution, you need two linearly independent solutions, right?

Only one is known and other one is unknown right?

Can you remember the method of reduction or order in this case?

For the second unknown solution, the reduction of order suggests thus,

try a second solution in the form of unknown function g(x) times and

y1, y1is equal to e to the x, u is unknown, okay?

2:28

That's the method of reduction of order, okay?

Plug in this expression into that,

what I mean is just to compute x y2 double

prime- 1 + x y2 prime + .y2,

compute it and simplify it?

Compute it xy double prime- 1 + x y2 prime + y2, okay?

Using this simplify then you

will get e to x times x u'' +

x- 1 u prime = 0, okay?

So that means inside the bracket must be equal to zero.

In other word x u double prime + (x-

1) u prime must be = 0, right?

Set v = u prime, then this equation becomes first order, okay?

Say xv prime + (x- 1)v =0,

can you see it, right?

xv prime now we have +(x -1) v = 0.

This is very simple, variable coefficient,

first order homogeneous differential equation, okay?

You can solve it in a couple of different ways, okay?

Maybe you can find this instigating factor.

Or you can recognize it as a separable differential

equations because it's the same as this given

differential equation is the same as dv,

over v + x -1 over x dx=0.

This is a separable differential equation,

which is very easy to solve, right?

4:40

So with this as our probable first order differential equation what you'll get is,

that v, v is in fact a u prime, that is equal to,

some arbitrary constant c sub 1 times x times e to the -x, right?

So what is u?

u is antiderivative of this expression so

you're going to get u is equal to arbitrary

constant c1 times -x e to the -x- e to the -x +

another arbitrary constant c sub 2, right?

5:24

We do not need the whole family of u right?

We do not need the whole family of u, okay?

We need just the nonconstant to function u so

that y2 becomes linearly independent from y1, okay?

So for example, we can take say c sub 1 = -1,

c sub 1 to be -1, and the c sub 2 = 0, okay?

5:54

By this choice, u becomes (x+1) times e to the -x, right?

And that means y2 becomes y2 = e to

the x times u that is equal to x + 1,okay?

So now you have for this given the nonhomogeneous equation.

We now have this is equal to, our competition shows this is = (x + 1), okay?

If you wish, you can check that this function x+1 is also

a solution of this homogeneous different equation, okay?

Moreover, this y1 e to the x and

this y to that is x + 1 they are linearly independent, okay?

So they form a fundamental set of solutions to this

homogeneous second order differential equation.

So in other words, the general solution of the corresponding

homogeneous problem is c1e to the x + c2(x+1), okay?

7:11

What are their [INAUDIBLE], okay?

You can compute the [INAUDIBLE] of W.

Which is [INAUDIBLE] (y1, y2), through the simple

computation you can see that is = -x e to the x, right?

7:29

Now it's time to find the particular solution of the original

nonhomogeneous problem, okay?

Particular solution of a nonhomogenous problem, okay?

As the variation of parameters says, we try yp, which is u1 y1.

u1 is unknown and e to the x + u2 unknown times y2,

that is x+1, right?

We try particular solution y p of this original nonhomogeneous

differential equation in the form of u1 e to the x + u2 x + 1, okay?

8:20

The two unknown functions u1 and u2 they satisfy simultaneous equation.

Say y1 u1 prime + y2 u2 prime = 0 and

y1 prime u1 prime + y2 prime

u2 prime = e of x, right?

Let's go back to our original differential equation, original problem, okay?

Our original problem is, and let me write it, okay?

This is equal to here is the input function

where originally this is the nonhomogeneous problem e to the x, right?

So when we are trying to find the particle

solution of this type u1 and

u2 must satisfy you know that u1 prime

= -g of y2 over capital W u2 prime = g

over y1 over capital W, right?

Can you recognize what should be g, okay?

For the g let's go back to the couple of slides, okay?

What is g, okay?

g is down there, g must be the right hand side side of

the second of the differential equation when the leading

coefficient is equal to 1 so, be careful a little bit.

From this equation, the given differential equation,

you should read it as y double prime, okay?

Let me write it here, right?

Let me write it there.

10:15

Yeah, this one, you should read it as,

y double prime- 1 + x over x y prime +

1 over xy and that is = x times e2x.

Not actually squared bout x times it, okay?

10:53

So by solving, the second equation for

the unknown is y1 prime u1 prime +

y2 prime u2 prime = g(x).

g(x) is equal to x times e to the 2x, okay?

11:27

On the other hand, the u2 prime from this expression,

you will get u2 prime = -e to the 2x, right?

So what is u1?

Enter derivative of (x + 1) e to the x, okay?

Say u1 is equal to x times e to the x.

By choosing integrating constant it will be 0, okay?

On the other hand, u2 is an anti derivative of -e to the 2x so

you have u2 is = negative one-half e 2x by

taking the interior constant to be 0, okay?

And then this, the particular solution y sub p is equal to,

x times e to the x times e to the x- one-half e to the 2x(x + 1), okay?

In other words that is equal one-half times (x- 1) times e to the 2x, okay?

So, all together here we have general solution

of the corresponding homogeneous problem so

called the complementary solution.

So y is equal to that complementary solution c1e to the x + c2 (x + 1) and

plus a particular solution which is one-half (x- 1) times e to the 2x.