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In order to confirm the method of reduction of order,

Â let's consider the following example.

Â Knowing that e to the x is a solution of

Â xy double prime minus (x+1) y_prime + y = 0.

Â So xy double prime minus (x+1)

Â y_prime + y = 2 on the interval from 0 to infinity.

Â First note that the the equation

Â that we are going to solve is nonhomogeneous linear differential equation.

Â For such a nonhomogeneous problem,

Â we know from the general theory its general solution will be y = y_c+y_p.

Â In symbol what is the y of c?

Â Y of c is the so called complementary solution which is the general solution of

Â corresponding homogeneous equation and

Â y_sub_p is any one particular solution of this nonhomogeneous equation.

Â So first we need to solve

Â the corresponding homogeneous problem which is

Â reduced to finding two linear independent solution of this differential equation.

Â Among these two, one is known as e to the x.

Â So we need to find another solution of

Â this homogeneous equation which is linearly independent of e to the x.

Â So let's set e to the x is equal to y_1.

Â Y_sub_1 is equal to e to the x by the reduction of order.

Â Set another solution y_sub_2 which we are assuming that,

Â which we wish to be linearly independent from y_sub_1 of

Â the form of y2 = u of x times e to the x where the u(x) is unknown.

Â Plugging this expression y_2 into the corresponding homogeneous equation,

Â into that equation, into this homogeneous equation.

Â And to simplify that expression then you will get xy

Â to double prime minus (x+1) y_2 prime plus y_2.

Â Compute these things and reorganize it,

Â then you are going to get e to the x times x u double prime plus x minus

Â 1 u prime is equal to zero.

Â That means that u must be a solution of

Â xu double prime plus x minus 1 u prime equal to zero.

Â Or rather by setting v is equal to u prime,

Â we are supposed to solve x v_prime plus (x-1)v =0.

Â So we need to solve here x

Â v_prime plus (x-1) and that is equal to zero.

Â That's a very simple first-order homogeneous differential equation

Â in v. Dividing through by x,

Â you are going to get v_prime plus 1-1/x times v is equal to 0

Â then it's the so-called integrating factor v of x is equal to e to

Â the integral of 1-1/x and d of x.

Â This is e to the x over x. Can you see it?

Â That's the integrating factor.

Â And so that means that if you multiply this v of x on this equation,

Â then it'll be e to the x over x times v

Â and the whole thing derivative and that is equal to zero.

Â So that means the e to the x over x and times v of x,

Â this is equal to constant c1.

Â So finally you get that v of x is equal to c1 x times e to the negative x.

Â But this v is

Â equal to in fact this is u prime.

Â So through integration again,

Â let's try to solve for u then.

Â U is equal to integral of this,

Â antiderivative of this function plus arbitrary integrating constant which is the same

Â as u of x is equal to negative c1 times x+1 and e to the negative x plus c_sub_2.

Â Because we need a non-constant function u of x,

Â so we may take c_2 is equal to zero and we may take c_1 is equal to -1.

Â Then our u of x becomes x+1 times e to the negative x.

Â So what is y_2 then?

Â Y_2 is equal to e to the x times u of x.

Â Our choice of e of x is x+1 times e to the negative x

Â so that now the y_2 becomes x+1.

Â So this y_2 x+1 is another solution of this homogeneous differential equation.

Â Moreover, this y_2 is independent of y_sub_1 which is equal to e to the x.

Â And that means the complementary solution say

Â y_sub_c is given by the c_1 times e to the x plus c_2 times x+1,

Â where the c_1 and the c_2 are arbitrary two constants.

Â This is the general solution of corresponding homogeneous differential equation.

Â So now we know completely what is y of c. Now only thing remaining is to

Â find one particular solution of original nonhomogenous differential equation.

Â But careful inspection shows us that y_p is equal 2.

Â Any constant, its derivatives, second derivative vanishes.

Â So if we choose y_p is equal 2 then that's

Â a trivial solution of this nonhomogeneous problem.

Â So in other words,

Â through the simple inspection way,

Â we found the one particular solution y_p is equal 2.

Â So the general solution of

Â original nonhomogeneous differential equation will be y is equal to

Â c_1 times e to the x plus c_2 times x plus 1 and plus 2.

Â This is the set of all possible solutions for this nonhomogeneous differential equation.

Â