In order to confirm the method of reduction of order, let's consider the following example. Knowing that e to the x is a solution of xy double prime minus (x+1) y_prime + y = 0. So xy double prime minus (x+1) y_prime + y = 2 on the interval from 0 to infinity. First note that the the equation that we are going to solve is nonhomogeneous linear differential equation. For such a nonhomogeneous problem, we know from the general theory its general solution will be y = y_c+y_p. In symbol what is the y of c? Y of c is the so called complementary solution which is the general solution of corresponding homogeneous equation and y_sub_p is any one particular solution of this nonhomogeneous equation. So first we need to solve the corresponding homogeneous problem which is reduced to finding two linear independent solution of this differential equation. Among these two, one is known as e to the x. So we need to find another solution of this homogeneous equation which is linearly independent of e to the x. So let's set e to the x is equal to y_1. Y_sub_1 is equal to e to the x by the reduction of order. Set another solution y_sub_2 which we are assuming that, which we wish to be linearly independent from y_sub_1 of the form of y2 = u of x times e to the x where the u(x) is unknown. Plugging this expression y_2 into the corresponding homogeneous equation, into that equation, into this homogeneous equation. And to simplify that expression then you will get xy to double prime minus (x+1) y_2 prime plus y_2. Compute these things and reorganize it, then you are going to get e to the x times x u double prime plus x minus 1 u prime is equal to zero. That means that u must be a solution of xu double prime plus x minus 1 u prime equal to zero. Or rather by setting v is equal to u prime, we are supposed to solve x v_prime plus (x-1)v =0. So we need to solve here x v_prime plus (x-1) and that is equal to zero. That's a very simple first-order homogeneous differential equation in v. Dividing through by x, you are going to get v_prime plus 1-1/x times v is equal to 0 then it's the so-called integrating factor v of x is equal to e to the integral of 1-1/x and d of x. This is e to the x over x. Can you see it? That's the integrating factor. And so that means that if you multiply this v of x on this equation, then it'll be e to the x over x times v and the whole thing derivative and that is equal to zero. So that means the e to the x over x and times v of x, this is equal to constant c1. So finally you get that v of x is equal to c1 x times e to the negative x. But this v is equal to in fact this is u prime. So through integration again, let's try to solve for u then. U is equal to integral of this, antiderivative of this function plus arbitrary integrating constant which is the same as u of x is equal to negative c1 times x+1 and e to the negative x plus c_sub_2. Because we need a non-constant function u of x, so we may take c_2 is equal to zero and we may take c_1 is equal to -1. Then our u of x becomes x+1 times e to the negative x. So what is y_2 then? Y_2 is equal to e to the x times u of x. Our choice of e of x is x+1 times e to the negative x so that now the y_2 becomes x+1. So this y_2 x+1 is another solution of this homogeneous differential equation. Moreover, this y_2 is independent of y_sub_1 which is equal to e to the x. And that means the complementary solution say y_sub_c is given by the c_1 times e to the x plus c_2 times x+1, where the c_1 and the c_2 are arbitrary two constants. This is the general solution of corresponding homogeneous differential equation. So now we know completely what is y of c. Now only thing remaining is to find one particular solution of original nonhomogenous differential equation. But careful inspection shows us that y_p is equal 2. Any constant, its derivatives, second derivative vanishes. So if we choose y_p is equal 2 then that's a trivial solution of this nonhomogeneous problem. So in other words, through the simple inspection way, we found the one particular solution y_p is equal 2. So the general solution of original nonhomogeneous differential equation will be y is equal to c_1 times e to the x plus c_2 times x plus 1 and plus 2. This is the set of all possible solutions for this nonhomogeneous differential equation.