The indicated function y1(x) is a solution of the given differential equation, the second solution is mathematically given as
[tex]y2=\frac{-1}{5x^2}[/tex]
Generally, the equation for the differential equation is mathematically given as
x^2y^2+2xy1-6y=0
Where
p(x)=2/x
[tex]y_2=y_1\inte\frac{p(x)dx}{y1^2}dx\\\\\int{1/y1^2e^{\intp(x)dx}}dx=\int1/x^4.xe^{-2/x dx}dx\\\\\int{1/y1^2e^{\intp(x)dx}}dx=\int1/x^4.xe^{-2logx}dx\\\\\int{1/y1^2e^{\intp(x)dx}}dx=\int1/x^4.x * x^{-2}dx[/tex]
y2=x^2*-1/5x^5=-1/5x^3
[tex]y2=\frac{-1}{5x^2}[/tex]
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