Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 13 and 14, use variation of parameters to find a general solution to the given equation for x>0.

x²y"(x)+2xy'(x)-2y(x)=6x^-2+3x

Step-by-Step Solution

Answer

By using the parameters, the General solution of the given equation is y(x) =c₁x+c₂x^-2 -9/14 x^-6 -9/4 x^-3 .

Definition of Cauchy-Euler equation:

It has direct application to Fourier's approach in the study of partial differential equations, the Cauchy-Euler equation is significant in the theory of linear differential equations.

Find the homogeneous solution:

The given differential equation is,

x²y"(x)+2xy'(x)-2y(x)=6x^-2+3x

In order to obtain solution via variation of parameter approach, first we need two linearly independent solutions of the humogen’s equation

x²y"(x)+2xy'(x)-2y(x)=0

Since x=0 is not an ordinary point, therefore we will use Cauchy-Euler equi-dimensional approach to find the solution of the equation.

Let  

L[y](x) = x²y"+2xy'-2y

And w(r,x) = x^r be the solution of the equation. Thus substituting w in the operator L, you get,

L[w](x) = x²(x^r )"(x)+2x(x^r )'-2(x^r )

=x² (r(r-1)) x^r-2 +2x(r) x^r-1 -2x^r

=r(r-1) x^r +2rx^r-2x^r

=(r²+r-2) x^r

Now solving the indicial equation is r²+r-2,

r²+r-2 =0

(r-1)(r+2)=0

r=1, -2

Thus, the two linearly independent solutions of the above homogeneous equation are y₁=x and y₂=x^-2.

Calculate the wronskians equation:

The differential equation can also be written as,

y"+2/x y'-2/x² y=6x^-4+3x^-1

The complimentary solution of the above equation is

y_c= c₁x+c₂x^-2

Let the particular solution y_p=u₁y₁+u₂y₂,  where

u₁'=W₁/W  and  u₂'= W₂/W

And

The values of different parameters are,

f(x) = 6x^-4+3x^-1  ,  y₁ = x  , and y₂ = x^-2

Now calculating Wronskians,

W=2x²-x²

W=x²

=0 - x^-2 (6x^-4+3x^-1)

= -6x^-6 - 3x^-3

=x (6x^-6 - 3x^-1) -0

=6x^-3+3

Find the general solution:

Calculating u₁:

u₁'= ( -6x^-6 - 3x^-3 )/x²

u₁'= -6x^-8 - 3x^-5  

u₁ = ∫ ( -6x^-8 - 3x^-5) dx

u₁ = 6/7 x^-7 +3/4 x^-4

Similarly Calculating u₂.  

u₂'= (6x^-3 +3 )/x² 

u₂'= 6x^-5 +3x^-2

u₂ = -6/4 x^-4-3 x^-1

The Particular solution will be,

y_p=u₁y₁ + u₂y₂

= (6/7 x^-7 +3/4 x^-4) x - ( -6/4 x^-4-3 x^-1 ) x^-2

=6/7 x^-6 +3/4 x^-3 x + 6/4 x^-6+3 x^-3

=-9/14 x^-6 -9/4 x^-3

The general solution of the given differential equation is,

y(x) = y_c + y_p

=c₁x+c₂x^-2 -9/14 x^-6 -9/4 x^-3

There for by using the parameters, the General solution of the given equation is y(x) =c₁x+c₂x^-2 -9/14 x^-6 -9/4 x^-3 .