AngouriMath
SolveAnExponentialTimesAnOddHalfPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
An exponential of a linear times a polynomial times an odd half power of a
quadratic,e^(a x + b) P Q^(m/2) , closed by the ansatz F = e^(a x + b) R Q^(k + 1/2) with m = 2k - 1 : F' = e^(a x + b) Q^(k - 1/2) (a R Q + R' Q + (k + 1/2) R Q') ,
anda R Q + R' Q + (k + 1/2) R Q' = P is a linear system in the coefficients
ofR , exact.
quadratic,
and
of
Remarks
Timofeev's e^x (1 - x - x^2)/sqrt(1 - x^2) is (e^x sqrt(1 - x^2))' and
had no antiderivative: by parts goes round in a circle and nothing substitutes.
Liouville's theorem says the elementary antiderivative ofe^(a x) times an
algebraic function, where there is one, ise^(a x) times an algebraic function
of the same field, which is what is looked for; there is no remainder term, unlike
the reduction without the exponential, so an empty solution is a decline.
https://github.com/asc-community/AngouriMath/issues/718
had no antiderivative: by parts goes round in a circle and nothing substitutes.
Liouville's theorem says the elementary antiderivative of
algebraic function, where there is one, is
of the same field, which is what is looked for; there is no remainder term, unlike
the reduction without the exponential, so an empty solution is a decline.
https://github.com/asc-community/AngouriMath/issues/718
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