3.6 Method of Reduction of Order

 [asciimath]y''-(4x+4)/xy'+(4x+8)/xy = 0[/asciimath]

2. Identify [asciimath]p(x)[/asciimath], the function coefficient of [asciimath]y'[/asciimath] and then find [asciimath]mu(x)[/asciimath].

 [asciimath]p(x)=-(4(x+1))/x \ ->[/asciimath]   [asciimath]mu(x)=e^(-int p(x)dx)=e^(4int (x+1)/xdx)[/asciimath] [asciimath]=e^(4x+4ln(x))=x^4e^(4x)[/asciimath] 

3. The second solution is given by

  [asciimath]y_2= y_1 int (mu(x))/((y_1)^2) dx[/asciimath]

 [asciimath]y_2= e^(2x) int (x^4e^(4x))/((e^(2x))^2) dx[/asciimath]

 [asciimath]=e^(2x) int x^4 dx[/asciimath]

 [asciimath]=1/5x^5e^(2x)+C[/asciimath]

We are looking for the simplest [asciimath]y_2[/asciimath], so we let the constant of integration be zero. Given any scalar multiple of [asciimath]y_2[/asciimath] is also a solution, we can choose [asciimath]y_2=x^5e^(2x)[/asciimath] as the simplest second solution.

a. Finding the second solution of the complementary equation:

We follow the steps for the reduction of orders method to find the second linearly independent solution to the complementary equation.

1a. First, standardize the equation by dividing it by the coefficient of [asciimath]y''[/asciimath]:

 [asciimath]y''+x^-1y'-x^-2y = 0[/asciimath]

2a. Identify [asciimath]p(x)[/asciimath], the function coefficient of [asciimath]y'[/asciimath] and then find [asciimath]mu(x)[/asciimath].

 [asciimath]p(x)=x^-1 \ ->[/asciimath]  [asciimath]mu(x)=e^(-int p(x)dx)=e^(-int x^-1dx)[/asciimath] [asciimath]=e^(ln(x))=x^-1[/asciimath] 

3a. The second solution is given by

  [asciimath]y_2= y_1 int (mu(x))/((y_1)^2) dx[/asciimath]

 [asciimath]y_2= x int (x^-1)/(x^2) dx[/asciimath]

 [asciimath]=x int x^-3 dx[/asciimath]

 [asciimath]=-1/2x^-1+cancel(C)^0[/asciimath]

 4a. The general solution to the complementary equation is

[asciimath]y_c=c_1y_1+c_2y_2[/asciimath]

 [asciimath]=c_1x+c_2(-1/2x^-1)[/asciimath]

Constant [asciimath]c_2[/asciimath] ​can absorb any numerical coefficients of  [asciimath]y_2.[/asciimath]  Thus [asciimath]y_c[/asciimath] simplifies to

 [asciimath]y_c=c_1x+c_2x^-1[/asciimath]

b. Finding a particular solution of the nonhomogeneous equation:

We use the method of variation of parameters to find the particular solution [asciimath]y_p[/asciimath].

1b. Standardize the original differential equation.

 [asciimath]y''+x^-1y'-x^-2y = x^-2(x^2+1)[/asciimath]

2b. The solutions to the homogeneous equation are now known: [asciimath]y_1=x[/asciimath] and [asciimath]y_2=x^-1[/asciimath].

The Wronskian of the fundamental set is

 [asciimath]W(y_1,y_2) = |(y_1,y_2),(y_1',y_2')|[/asciimath]

 [asciimath]W (x)= |(x,x^-1 ),(1,-x^-2) |[/asciimath]

 [asciimath]=-2x^-1[/asciimath]

3b. Next substituting [asciimath]y_1=x[/asciimath]  , [asciimath]y_2=x^-1[/asciimath], [asciimath]f(x)=x^-2(x^2+1)[/asciimath], and [asciimath]W(y_1,y_2)=-2x^-1[/asciimath] into formulas for [asciimath]u_1[/asciimath] and  [asciimath]u_2[/asciimath] to determine them.

Finding [asciimath]u_1[/asciimath]:

    [asciimath]u_1=int (-f(x) y_2)/(W(y_1,y_2)) dx[/asciimath]

 [asciimath]u_1=int (-x^-2(x^2+1) x^-1)/(-2x^-1) dx[/asciimath]

[asciimath]=1/2intx^-2(x^2+1)dx[/asciimath]

[asciimath]=1/2int(1+x^-2)dx[/asciimath]

 [asciimath]=1/2(x-1/x)[/asciimath]

Finding [asciimath]u_2[/asciimath]:

    [asciimath]u_2=int (f(x) y_1)/(W(y_1,y_2)) dx[/asciimath]

 [asciimath]u_2=int (x^-2(x^2+1) x)/(-2x^-1) dx[/asciimath]

 [asciimath]=-1/2int(x^2+1)dx[/asciimath]

 [asciimath]=-1/2(x^3/3+x)[/asciimath]

4b. Substitute [asciimath]u_1[/asciimath] and [asciimath]u_2[/asciimath] together with [asciimath]{y_1,y_2}[/asciimath] into the expression for [asciimath]y_p[/asciimath] to obtain a particular solution:

 [asciimath]y_p = u_1 y_1 + u_2 y_2[/asciimath]

 [asciimath]y_p=1/2(x-1/x) x-1/2(x^3/3+x) x^-1[/asciimath]

 [asciimath]=1/3x^2-1[/asciimath]

 c. Finding the General Solution

 The general solution to the nonhomogeneous equation is the sum of the particular solution and complementary solution.

 [asciimath]y=y_p+y_c[/asciimath]

 [asciimath]y(x)=1/3x^2-1+c_1x+c_2x^-1[/asciimath]

d. Applying the initial conditions

Applying the initial condition to [asciimath]y[/asciimath]:

 [asciimath]y(1)=2[/asciimath]

 [asciimath]1/3(1^2)-1+c_1(1)+c_2(1^-1)=2[/asciimath]

 [asciimath]1/3-1+c_1+c_2=2[/asciimath]

[asciimath]c_1+c_2=8/3[/asciimath]

Applying the initial condition to [asciimath]y'[/asciimath]:

 [asciimath]y'(x)=2/3x+c_1-c_2x^-2[/asciimath]

 [asciimath]y'(1)=-3[/asciimath]

 [asciimath]2/3(1)+c_1-c_2(1^-2) =-3[/asciimath]

[asciimath]c_1-c_2=-11/3[/asciimath]

To determine [asciimath]c_1[/asciimath]  and [asciimath]c_2[/asciimath], we solve the following system of two equations and two unknowns:

 [asciimath]{(c_1+c_2=8/3),(c_1-c_2=-11/3 ):}[/asciimath]

Solving the system yields

 [asciimath]c_1=-1/2,[/asciimath]    [asciimath]c_2=19/6[/asciimath]

Therefore the solution to the initial value problem is

 [asciimath]y(x)=1/3x^2-1-1/2x+19/6x^-1[/asciimath]