Hi there. I have this differential equation: [tex]x^4y''+2x^3y'-y=0[/tex] And I have to find one solution of the form: [tex]\sum_0^{\infty}a_nx^{-n},x>0[/tex] So I have: [tex]y(x)=\sum_0^{\infty}a_n x^{-n}[/tex] [tex]y'(x)=\sum_1^{\infty}(-n) a_n x^{-n-1}[/tex] [tex]y''(x)=\sum_2^{\infty}(-n)(-n-1) a_n x^{-n-2}[/tex] Then, replacing in the diff. eq. [tex]x^4\sum_2^{\infty}(-n)(-n-1) a_n x^{-n-2}+2x^3\sum_1^{\infty}(-n) a_n x^{-n-1}-\sum_0^{\infty}a_n x^{-n}=0[/tex] [tex]\sum_2^{\infty}(-n)(-n-1) a_n x^{-n+2}+2\sum_1^{\infty}(-n) a_n x^{-n+2}-\sum_0^{\infty}a_n x^{-n}=0[/tex] Expanding the first term for the second summation, and using k=n-2->n=k+2 for the first and the second sum: [tex]-\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x-2\sum_0^{\infty}(k+2) a_n x^{-k}-\sum_0^{\infty}a_k x^{-k}=0[/tex] [tex]-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+4) a_{k+2} +a_k \right ]=0[/tex] Then: [tex]-2a_1=0\rightarrow a_1=0[/tex] [tex]a_{k+2}=\frac{-a_k}{(k+2)(k+4)}[/tex] After trying some terms I get for the recurrence relation: [tex]a_{2n}=\frac{(-1)^n a_0}{2^{2n}(n+1)!n!}[/tex] And [tex]a_{2n+1}=0\forall n[/tex] So then I have one solution: [tex]y(x)=\sum_{n=0}^{\infty}\frac{(-1)^n a_0}{2^{2n}(n+1)!n!}x^{-2n}[/tex] Now, I think this is wrong, but I don't know where I've committed the mistake. I hoped to find a series expansion for [tex]cosh(1/x)[/tex] or [tex]sinh(1/x)[/tex] because wolframalpha gives the solution: [tex]y(x)=c_1 cosh(1/x)-ic_2 sinh(1/x)[/tex] For the differential equation (you can check it here) Would you help me to find the mistake in here?
Ok. I've found a mistake there. [tex]-\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x \color{red} - \color{red} 2\sum_0^{\infty}(k+2) a_n x^{-k} -\sum_0^{\infty}a_k x^{-k}=0[/tex] So this gives: [tex]-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+5) a_{k+2} +a_k \right ]=0[/tex] And the recurrence formula: [tex]a_{k+2}=\frac{-a_k}{(k+2)(k+5)}[/tex] It's worse now, because I couldn't even find the recurrence relation.
Hi Telemachus! Always check the minuses first … should be … [tex]\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x-2\sum_0^{\infty}(k+2) a_n x^{-k}-\sum_0^{\infty}a_k x^{-k}=0[/tex] [tex]-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+1) a_{k+2} +a_k \right ]=0[/tex]