Q13E - 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)=x^-1/2

Step-by-Step Solution

Answer

By using the parameters, the General solution of the given equation is, 

y(x)=c₁x+c₂x²+(4/15) x^-1/2

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)=x^-1/2

In order to obtain a 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 equidimensional approach to find the solution of the equation.

Let 

L[y](x)=x²y"-2xy'+2y

And 

w(r,x)=x^r

This will be the solution of the equation. Thus substituting w in the operator L, you get

L[y](x)=x²(x^r)"-2x(x^r)'+2(x^r)

=x² [r(r-1)]x^r-2 -2xr(x)^r-1+2x^r

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

=(r²-3r+2) x^r

Now solving the indicial equation is,

r²-3r+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=x².

Calculate the Wronskians equation:

The differential equation can also be written as,

y"-(2/x) y'+(2/x²)y=x^-5/2

The complimentary solution of the above equation is,

u₁'=W₁/W  and  u₂'=W₂/W

W= |y₁   y₂|

      |y'₁   y'₂| 

W₁= |0   y₂|

      |f(x)   y'₂| 

W₂= |y₁   0|

      |y'₁   f(x)|

The values of different parameters are,

f(x)=x^-5/2  , y₁=x   ; y₂=x²

Now Calculating Wronskians,

W=|x   x²|

     |1   2x|

=2x²-x²

=x²

W₁=|0       x²|

       |x^-5/2  2x|

=0-x²-x^-5/2

=-x^-1/2

W₂ =|x      0|

       |1  x^-5/2 |

=(x) x^-5/2-0

=-x^-3/2

Find the general solution:

Calculating u₁:

u₁'= -x^-1/2/x²

= -x^--5/2

u₁=∫  -x^--5/2 dx

=2/3 x^-3/2

Similarly Calculating u₂.

u₂'= x^-3/2/x²

= x^--7/2

u₂=∫x^--7/2 dx

= -2/5 x^-5/2

The Particular solution will be,

y=u₁y₁+u₂y₂

=2/3 x^-3/2 x- 2/5 x^-5/2 x²

=2/3 x^-1/2 - 2/5 x^-1/2  

=4/15 x^-1/2

The general solution of the given differential equation is,

y(x)=y_c+y_p

=c₁x+c₂x²+(4/15) x^-1/2