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Cuestion 2 EDiferenciales 1109S2: Page 7 of 7

Solapas principales

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\sum_{n=0}^{\infty}( (2\cdot k\cdot n+2\cdot k)\cdot a_n + (k\cdot x-k\cdot 4)\cdot a_n +  
+(n+1)\cdot (n)\cdot b_n+ (x\cdot (n+1)-
-3\cdot (n+1))\cdot b_n +  3 \cdot b_n )\cdot x^{n+1}=0
 
\sum_{n=0}^{\infty}( (2\cdot k\cdot n)\cdot a_n + (k\cdot x-k\cdot 2)\cdot a_n +(n+1)\cdot (n)\cdot b_n+ (x\cdot (n+1)+ (-3\cdot n-3\cdot))\cdot b_n +  3 \cdot b_n )\cdot x^{n+1}=0
 
\sum_{n=0}^{\infty}( (2\cdot k\cdot n)\cdot a_n + (k\cdot x-k\cdot 2)\cdot a_n +(n+1)\cdot (n)\cdot b_n+ (x\cdot (n+1)+ (-3\cdot n))\cdot b_n )\cdot x^{n+1}=0
 
(2\cdot k\cdot n)\cdot a_n + (k\cdot x-k\cdot 2)\cdot a_n +(n+1)\cdot (n)\cdot b_n+ (x\cdot (n+1)+ (-3\cdot n))\cdot b_n=0
 
 
(2\cdot k\cdot n)\cdot a_n + (k\cdot x-k\cdot 2)\cdot a_n +(n^2+n)\cdot b_n+ (x\cdot (n+1)+ (-3\cdot n))\cdot b_n=0
 
b_n=\frac{2\cdot k\cdot n+ (k\cdot x-k\cdot 2)}{(n^2+n)+(x\cdot (n+1)+ (-3\cdot n))}\cdot a_n
 
y_2(x)=k\cdot y_1(x)\cdot ln(x)+\sum_{n=0}^{\infty} \frac{2\cdot k\cdot n+ (k\cdot x-k\cdot 2)}{(n^2+n)+(x\cdot (n+1)+ (-3\cdot n))}\cdot a_n \cdot x^{n+r}=
=\sum_{n=0}^{\infty} k\cdot (ln(x)+\frac{2\cdot n+ ( x- 2)}{(n^2+n)+(x\cdot (n+1)+ (-3\cdot n))})\cdot a_n=
 
=\sum_{n=0}^{\infty} k\cdot (ln(x)+\frac{2\cdot n+ ( x- 2)}{(n^2+n)+(x\cdot (n+1)+ (-3\cdot n))})\cdot a_n=
=-\sum_{n=0}^{\infty} k\cdot (ln(x)+\frac{2\cdot n+ ( x- 2)}{(n^2+n)+(x\cdot (n+1)+ (-3\cdot n))})\cdot\frac{(n+1)!}{(n!\cdot (n-2)!)}\cdot a_{0}
 
La solucion general de la ecuacion diferencial viene dada por:
 
y(x)=C_1\cdot y_1(x)+C_2\cdot y_2(x)
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