Find approximate solutions to  `x^3y +xy - y =0`  by substituting `x = e^t`

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We have `x^3y + xy-y = 0`

A trivial solution is `y=0``x in RR`

Otherwise we need to solve

`x^3 + x - 1 =0`

Draw the graph


The only real (`x in RR`) solution to this is positive,

therefore we can let  `x = e^t` , which is always positive.

Use a Taylor series approximation to the function `f(t) = e^(3t) + e^t -1`

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