xy'-y =sqrt((x^2)+y(^2))(adsbygoogle = window.adsbygoogle || []).push({});

so by method of homogeneous equations I get arcsinh(y/x) = ln|x|+c but don't know what to do from there. the book suggests using sinhx = 1/2 (e^x - e^-x) but I don't know how to incorporate this. The answer should be y= (1/2) * (cx^2 - 1/c). How do you get this????

This is my work so far:

y'-(y/x) = sqrt (1+(y/x)^2)

since y=ux...

du/(sqrt(1+u^2) = dx/x

taking the antiderivatives gives

arcsinh(u) = ln|x|+c

where

arcsinh|y/x| = ln|x|+c

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# Homework Help: Basic differential equations, homogeneous method

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