So, fortunately we have two equations

where the two unknowns are u_1 and u_2, right?

What are they?

First one is this one.

Require u_1'y_1'+u_2'y_2'=0.

This is one such equation.

Now, we have another equation for the unknowns u_1 and u_2, right here.

u_1'y_1'+u_2'y_2'=0.

We have two unknowns,

u_1 and u_2, and we have two equations,

u_1'y_1+u_2'y_2=0

and u_1'y_1'+u_2'y_2'=g.

That's what we are expecting in fact.

What I mean is, therefore,

we obtain a system of equations with two unknowns,

u_1 prime and u_2 prime.

Say, y_1u_1'+y_2u_2'=0.

Another one is, y_1'u_1'+y_2'u_2'=g.

We have two unknowns,

u_1_prime and u_2_prime and we have two equations.

This is a system of equations for two unknowns and linear,

a linear system of equations for two unknowns, u_1_prime and u_2_prime.

Apply the Cramer's rule.

By the Cramer's rule for solving this linear system of equation then,

first unknown u_1 prime is equal

to determinant of the coefficient matrix.

What is the determinant of the coefficient matrix?

Coefficient matrix is y_1,

y_2, and y_1 prime, and y_2 prime.

We need this determinant, right?

Can you remind what it is?

We use this one as a W.

This is the so called the Wronskian of two linearly independent functions y_1 and y_2,

where W of y_1 and y_2,

which I simply write by capital W.

Its determinant is equal to y_1,

y_2 prime minus y_2, y_1 prime.

So, by the Cramer's rule,

the first unknown u_1_prime is equal to determinant of two by

two matrix over determinant of the coefficient matrix which is the Wronskian.

So, in fact, this will be u_1_prime is

equal to minus y_2 times G over W. On the other hand,

the second unknown u_2_prime is equal to determinant of y_1 zero,

y_1_prime g over the Wronskian,

which is equal to y_1 times g over W.

So, in a sense, we solve the two unknowns, u_1_prime and u_2_prime.

u_1 and u_2, we can obtain by solving

this simple first-order differential equation or simply through the integration.

What is u_1?

u_1 must be antiderivative of this part.

And what is u_2?

u_2 must be the antiderivative of this expression.

Obtaining u_1 and u_2 in this way,

due to this u_1 and u_2,

and make the final Y_P.

This is u_1, y_1 plus u_2, y_2,.

y_1 and y_2, they are given or they are known already;

u_1 and u_2, you can obtain by solving by integrating the equation four.

Using those four functions for y of p,

you get a particular solution of the original problem.

So, the general solution of the original problem,

general solution to y double prime

plus p y_prime plus qy is equal to g. Now,

you can get y is equal to, y_c is what?

c_1 y_1 plus c_2 y_2 and plus y_p.

This is y of c, complementary solution.

y of p, that is equal to u_1 y_1 plus u_2 y_2.

This is a particular solution y of p.

So, we are done.

We call this process of obtaining a particular solution of

this form for the nonhomogeneous differential equation,

we call it the variation of parameters

because you are replacing arbitrary constant c_1 and c_2 by unknown function u_1 and u_2.

In summary, we can state,

we can have the following theorem,

the variation of parameters.

Assume that all the functions involved,

say, p of x and q of x and g of x,

they are continuous in certain interval from alpha to beta,

and assume that y_1 and y_2

are two linearly independent solutions of corresponding homogeneous equation,

say, y double prime plus p(x) y_prime plus q(x) y is equal to zero,

then, the particular solution is given by y_1 times u_1.

What is u_1?

Antiderivative of negative y_2 g over W plus y_2 times u_2.

What is u_2?

Antiderivative of y_1 g over W. Can you remind it?

I'll show it over here.

u_1 prime is equal to negative y_2 g over W.

On the other hand,

u_2 prime is equal to y_1 g over W.

So, u_1 is an antiderivative of negative y_2 g over W and

u_2 is antiderivative of y_1 g over W. That's what I'm writing here.

So, y_1 times u_1 and y_2 times this is u_2 part,

where the capital W,

this is Wronskian of two linearly independent function y_1 and y_2.

So this is the required particular solution of our original homogeneous problem.

So finally, as I said here,

the general solution of our original problem is y_c plus y_p,

and y_c is simply linear combination of y_1 and the y_2.

So, we completely solve

this possibly variable coefficient linear nonhomogeneous differential equation.

I'll illustrate this theorem,

the variation of parameters, through a couple of examples.