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implicit differentiation y-x=y(x+y) find y''
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y' - 1 = y'(x+y)+(1+y')(y) y'-1 = xy'+yy' +y+yy' y' -xy'-yy'-yy' = y+1 y'(1-x-2y) = y+1 solve for y' should be \[y' =\frac{ y+1 }{ 1-x-2y }\]
how about y''?
take d/dx of y'
\[y" =\frac{ y'(1-x-2y)-(y+1)(-1-2y') }{(1-x-2y)^2 }\] wherever you see a y' plug equation u got for y' it's a really messy problem...
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