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`
Now, using standard Taylor series approximations
`e^(3t) approx 1 + 3t + (3t)^2/2` and
`e^t approx 1 + t + t^2/2`
Therefore `e^(3t) + e^t -1 approx 2 + 4t + (10t^2)/2 -1 = 1 +4t + 5t^2`
We now want to solve
`5t^2 + 4t + 1 approx 0`
Using the quadratic formula
`implies` `t approx (-4 +- sqrt(16 - 20))/10 = -2/5 + r`
If the solution is real (it must be to insure `x` is real) and unique then `r approx 0`
and we have that `t approx -2/5`
`implies` `x approx e^(-2/5) = 0.67`
Check
`x^3 + x -1 approx 0.67^3 +0.67 -1 = -0.028`
which is approximately zero.
Therefore ` `the possible solutions are
`y=0`, `x in RR` or `x approx 0.67`,`y in RR`
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