To make the thing simple,

we consider only second order non-homogeneous differential equation.

So consider non-homogeneous linear second order differential equation with

possibly variable coefficients in its standard form y" + p(x)y' + q(x)y = g(x), right?

Assume that we have

two linear independent solution of corresponding homogeneous problem.

Corresponding homogeneous problem,

trivially that is equal to y" + p(x)y' + q(x)y = 0, right?

This is the corresponding homogeneous problem.

We assume that y_1 and y_2 are

two linear independent solutions of this homogeneous problem.

In other words, we have a complementary solution y_c = c_1. y_1 + c_2.y_2, right?

And the general solution of this reason given

non-homogeneous question will be the general solution of

differential equation one is y

is equal to complementary solution plus particular solution.

We already know what is this complementary solution y_c, right?

So only thing unknown is y_p now, okay?

The main idea of the method of variation

of parameters is to assume that there is

a particular solution of

this non-homogeneous problem in the form of y_p = u_1.y_1 + u_2.y_2, okay?

Where the u_1(x) and

the u_2(x) are two unknown functions to be determined, okay?

So that y_p becomes a particular solution of this original non-homogeneous problem.

Here in our guess for y_p, we've two unknowns the u_1 and u_2, okay?

So in order to determine two unknown functions,

the u_1 and the u_2,

we need two conditions.

Our guess is, you are looking for a particular solution of y_sub_p of

the given differential equation in the form of

y_p is equal to u_1 times y_1 plus, u_2 times y_2.

Since we have two unknowns,

as I said, u_1 and u_2,

we need two conditions to determine u_1 and u_2.

One condition trivially comes from

the condition that y_p is a particular solution of the given differential equation.

In other words, y_p double prime plus py_p prime,

plus qy_p is equal to g. Using this expression,

this equation will give us one equation for two unknowns,

u_1 and u_2, trivially.

We need one more equation.

That another equation, we will take it so that our next computation is to be easy.

Differentiating two.

Two is this.

This is the equation two.

Differentiating this y_p, you will get derivative of y_p will

be u_1 y_1 prime plus u_2 y_2 prime,

plus u_1 prime y_1,

and plus u_2 prime y_2.

It's easy by the product rule of differentiation.

From this expression, we require the second part to be equal to zero.

Say, u_1 prime y_1,

plus u_2 prime y_2,

to be equal to 0.

We simply require it without a proper reason at this moment.

Then, y_p prime will be substituted to prime 1.

In other words, u_1 y_1 prime plus u_2 y_2 prime.

So now, we have right away, from y_p prime,

this is equal to u_1 y_1 prime plus u_2 y_2 prime,

this is of course to plot,

plus u_1 prime y_1,

plus u_2 prime y_2.

We require this is equal to zero.

We simply require this is equal to zero.

Now what does that mean then?

That means, y_p is equal to only this part.

So, is a second derivative will be,

u_1 y_1 double prime,

u_2 y _2 double prime and plus u_1 prime y_1 prime,

plus u_2 prime y_2 prime,

by the product rule again.

So using this y_p,

and y_p prime, and y_p double prime,

make the left hand side of the differential equations, say,

y_p double prime plus p of x times

y_p prime plus true x of y

over p. Here's the computation,

y_p double prime plus p_y p prime,

plus q_y p. If we reorganize this sum of these three times then,

you are going to get u_1 times y_1 double prime plus p_y1 prime, plus q_y1,

plus u_2 times y_2 double prime, plus p_y2 prime,

plus q_y2, plus u_1 prime y_1 prime,

plus u_2 prime and y_2 prime.

Let's pay attention to the quantities in this bracket,

and the quantities in this bracket.

What can you say about these two things?

What I mean is y_1 double prime plus p_y1 prime plus

q_y1 and y_2 double prime plus p_y2 prime plus q of y_2.

What can you say about these two?

Can you remind what are those y_1 and y_2?

I said,

y_1 and y_2,

this is two linearly independent solutions

to corresponding homogeneous problem

y double prime plus p_x y prime,

plus true x y equal to zero.

In other words, y_1 double prime plus p_y1 prime, plus q_y1,

or y_2 double prime plus p_y2 prime plus q_y2,

both of them must be equal to 0,

because y_1 and y_2 are solutions of this corresponding homogeneous problem.

What does that mean?

The first bracket is equal to zero.

The second bracket, that is equal to zero.

It then means that in fact,

this term, this part,

and that part, they disappear.

So, we will get only u_1 prime times y_1 prime,

plus u_2 prime times y_2 prime because y_p,

this is a particular solution of our original problem.

Some of these three terms must be equal to g. Can you see why?

We have one more condition right here,

an unknown, the u_1 and the u_2.

We have a one on other such a condition, say,

u_1 prime y_1 plus u_2 prime y_2 is equal to 0.

So let me summarize those two equations here.