ecuació diferencial de producte integral invers

d_{xx}^{2}[ f(g(x)) [o(x)o] g(x)^{[o(x)o](-1)} ]  = (x^{m_{1}}+...(n)...+x^{m_{n}})·f(g(x))


f(g(x)) = e^{g(x)}
g(x) = int[ x^{n_{1}}+...(n)...+x^{m_{n}} ]d[x]

solució de ecuacions diferencials en series

d_{xx}^{2}[f(x)]=x^{n}·f(x)


f(x) = ∑ ( 1/((n+2)k+(n+1))(n+2)k+(n+2))! )·x^{(n+2)(k+1)}


d_{xx}^{2}[f(x)] = ∑ ( 1/((n+2)(k+(-1))+(n+1))((n+2)(k+(-1))+(n+2))! )·x^{(n+2)k}·x^{n}
d_{xx}^{2}[f(x)] = ∑ ( 1/((n+2)p+(n+1))((n+2)p+(n+2))! )·x^{(n+2)(p+1)}·x^{n}
k+(-1)=p & k=p+1


d_{x,...,x}^{m}[f(x)]=x^{n}·f(x)


f(x) = ∑ ( 1/((n+m)k+(n+1))·...·(n+m)k+(n+m))! )·x^{(n+m)(k+1)}