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As a last two topic in chapter four,

Â I'd like to introduce one more very strong way of

Â obtaining a particular solution for the linear non-homogeneous differential equation.

Â Though the method of undetermined coefficients is rather simple,

Â but it has two strong constraints.

Â Consider differential equation L|y| = g(x).

Â Non-homogeneous one, okay?

Â L is the linear differential operator.

Â L is the linear differential operator.

Â What is L?

Â Is equal to a_n(x) and D the to the n plus a_n - 1(x),

Â D to the n-1 and

Â plus a_1(x) and D and plus a_0(x), right?

Â That is L, differential operator of order n, okay?

Â So, consider such a differential equation, non-homogeneous differential equation.

Â For the method of undetermined coefficients to be applied,

Â first your differential operator L, okay,

Â must have only constant of coefficients.

Â In other words, all those coefficients, a_n, a_n-1, a_1,

Â and a_0, they must be a constant, okay?

Â None of them can be a true function of x.

Â Second constraint is coming

Â from the right-hand side of that differential equation L|y| = g(x),

Â g(x) some times we quote it as the input function, okay?

Â This input function must have a differential polynomial annihilator.

Â It must have differential

Â polynomial annihilator.

Â 4:14

To make the thing simple,

Â we consider only second order non-homogeneous differential equation.

Â So consider non-homogeneous linear second order differential equation with

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

Â Assume that we have

Â two linear independent solution of corresponding homogeneous problem.

Â Corresponding homogeneous problem,

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

Â This is the corresponding homogeneous problem.

Â We assume that y_1 and y_2 are

Â two linear independent solutions of this homogeneous problem.

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

Â And the general solution of this reason given

Â non-homogeneous question will be the general solution of

Â differential equation one is y

Â is equal to complementary solution plus particular solution.

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

Â So only thing unknown is y_p now, okay?

Â The main idea of the method of variation

Â of parameters is to assume that there is

Â a particular solution of

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

Â Where the u_1(x) and

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

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

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

Â So in order to determine two unknown functions,

Â the u_1 and the u_2,

Â we need two conditions.

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

Â the given differential equation in the form of

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

Â Since we have two unknowns,

Â as I said, u_1 and u_2,

Â we need two conditions to determine u_1 and u_2.

Â One condition trivially comes from

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

Â In other words, y_p double prime plus py_p prime,

Â plus qy_p is equal to g. Using this expression,

Â this equation will give us one equation for two unknowns,

Â u_1 and u_2, trivially.

Â We need one more equation.

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

Â Differentiating two.

Â Two is this.

Â This is the equation two.

Â Differentiating this y_p, you will get derivative of y_p will

Â be u_1 y_1 prime plus u_2 y_2 prime,

Â plus u_1 prime y_1,

Â and plus u_2 prime y_2.

Â It's easy by the product rule of differentiation.

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

Â Say, u_1 prime y_1,

Â plus u_2 prime y_2,

Â to be equal to 0.

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

Â Then, y_p prime will be substituted to prime 1.

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

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

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

Â this is of course to plot,

Â plus u_1 prime y_1,

Â plus u_2 prime y_2.

Â We require this is equal to zero.

Â We simply require this is equal to zero.

Â Now what does that mean then?

Â That means, y_p is equal to only this part.

Â So, is a second derivative will be,

Â u_1 y_1 double prime,

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

Â plus u_2 prime y_2 prime,

Â by the product rule again.

Â So using this y_p,

Â and y_p prime, and y_p double prime,

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

Â y_p double prime plus p of x times

Â y_p prime plus true x of y

Â over p. Here's the computation,

Â y_p double prime plus p_y p prime,

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

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

Â plus u_2 times y_2 double prime, plus p_y2 prime,

Â plus q_y2, plus u_1 prime y_1 prime,

Â plus u_2 prime and y_2 prime.

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

Â and the quantities in this bracket.

Â What can you say about these two things?

Â What I mean is y_1 double prime plus p_y1 prime plus

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

Â What can you say about these two?

Â Can you remind what are those y_1 and y_2?

Â I said,

Â y_1 and y_2,

Â this is two linearly independent solutions

Â to corresponding homogeneous problem

Â y double prime plus p_x y prime,

Â plus true x y equal to zero.

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

Â or y_2 double prime plus p_y2 prime plus q_y2,

Â both of them must be equal to 0,

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

Â What does that mean?

Â The first bracket is equal to zero.

Â The second bracket, that is equal to zero.

Â It then means that in fact,

Â this term, this part,

Â and that part, they disappear.

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

Â plus u_2 prime times y_2 prime because y_p,

Â this is a particular solution of our original problem.

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

Â We have one more condition right here,

Â an unknown, the u_1 and the u_2.

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

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

Â So let me summarize those two equations here.

Â