In 13 and 14, use variation of parameters to find a

Chapter 8, Problem 14E

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

$x^{2} y^{\prime \prime}(x)+2 x y^{\prime}(x)-2 y(x)=6 x^{-2}+3 x$

Equation Transcription:

$x>0$

${x}^{2}y''(x)+2xy'(x)-2y(x)=6{x}^{-2}+3x$

Text Transcription:

x>0

x2y''(x)+2xy'(x)-2y(x)=6x-2+3x

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