Given a second order linear homogeneous differential equation

a₂(x)y′′+a₁(x)y′+a₀(x)y=0

we know that a fundamental set for this ODE consists of a pair linearly independent solutions y₁,y₂. But there are times when only one function, call it y₁, is available and we would like to find a second linearly independent solution. We can find y₂ using the method of reduction of order.

First, under the necessary assumption the a₂(x)≠0 we rewrite the equation as

y′′+p(x)y′+q(x)y=0 p(x)=a₁(x)/a₂(x), q(x)=a0(x)/a2(x),

Then the method of reduction of order gives a second linearly independent solution as

y₂(x)=Cy₁u=Cy1(x) ∫ e−∫p(x)dx/y²₁(x)dx

where C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. Once useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain y₂=C3e²ˣ then we can choose C=1/3 so that y₂=e²ˣ.

Given the problem

x²y′′5xy−20y′+4y=0

and a solution y1=e(2x/5).

Applying the reduction of order method to this problem we obtain the following

y21(x)=_______