martes, 3 de septiembre de 2019

ecuació diferencial de lagranià

c·d_{x}[y(x)]^{n}+a(x)·( y(x) )^{p} = 0


y(x) = e^{f(x)}
ce^{n·f(x)}·d_{x}[f(x)]^{n}+a(x)·e^{p·f(x)} = 0
e^{p·f(x)}( ce^{(n+(-p))·f(x)}·d_{x}[f(x)]^{n}+a(x) ) = 0
ce^{(n+(-p))·f(x)}·d_{x}[f(x)]^{n}+a(x) = 0
c^{(1/n)}·e^{((n+(-p))/n)·f(x)}·d_{x}[f(x)] = (-a(x))^{(1/n)}
c^{(1/n)}·(n/(n+(-p)))·e^{((n+(-p))/n)·f(x)} = int[ (-a(x))^{(1/n)} ]d[x]
c^{(1/n)}·(n/(n+(-p)))·e^{((n+(-p))/n)·ln( y(x) )} = int[ (-a(x))^{(1/n)} ]d[x]
c^{(1/n)}·(n/(n+(-p)))·( y(x) )^{((n+(-p))/n)} = int[ (-a(x))^{(1/n)} ]d[x]


y(x) = ( (1/c^{(1/n)})·((n+(-p))/n)·int[ (-a(x))^{(1/n)} ]d[x] )^{(n/(n+(-p)))}




c·d_{x}[y(x)]^{n}+a(x)·( y(x) )^{n} = 0


y(x) = e^{f(x)}


ce^{n·f(x)}·d_{x}[f(x)]^{n}+a(x)·e^{n·f(x)} = 0
c^{(1/n)}·d_{x}[f(x)] = (-a(x))^{(1/n)}
c^{(1/n)}·f(x) = int[ (-a(x))^{(1/n)} ]d[x]
c^{(1/n)}·ln( y(x) ) = int[ (-a(x))^{(1/n)} ]d[x]


y(x) = e^{(1/c^{(1/n)})·int[(-a(x))^{(1/n)}]d[x]}







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