Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. x2y'' − 3xy' + 4y = 11x2 ln x, x > 0; y1(x) = x2, y2(x) = x2 ln x

Therefore the particular solution of the given differential equation is  

Step-by-step explanation:

The given ordinary differential equation is

If y₁(x) =x² is a solution of ODE then it will be satisfy the ODE

y₁'(x)= 2x and y₁"(x)=2

Putting the value of y₁'(x) and y₁"(x) in x²y"-3xy'+4y=0 we get

x².2-3x.2x+4x²= 2x²-6x²+4x²=0

Therefore y₁(x)  is a solution of the given differential equation.

Again,

y₂(x) =(x²ln x ) is a solution of ODE then it will be satisfy the ODE.

= 3+2 ln x

Putting the value of y₂'(x) and y₂"(x) in x²y"-3xy'+4y=0 we get

=0

Therefore  y₂(x) is a solution of  the given differential equation.

The wronskian of y₁(x) and  y₂(x)  is

            =x²(2x ln x+x)-x²ln x(2x)

            =2x³ ln x +x³ - 2x³ln x

            =x³≠0

Here

The particular solution is

                              Let ln x  =u  

Therefore the particular solution of the given differential equation is  

so, converting firstly the mixed fractions to improper fractions and then "adding".