Question #87056
Solve the following ODE using the power series method: (x^2-1) y′′ +3 xy′ + xy =0
1
Expert's answer
2019-04-03T10:05:57-0400

We seek the solution of the equation in the form of a power series


y(x)=n=0anxny(x)= \textstyle\sum_{n=0}^{\infty} a_n x^n

Then ODE:


(x21)n=2ann(n1)xn2+3xn=1annxn1+xn=0anxn=0(x^2-1)\sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} +3 x \sum_{n=1}^{\infty} a_n n x^{n-1} + x \sum_{n=0}^{\infty} a_n x^n = 0n=2ann(n1)xn2a26a3xn=2an+2(n+2)(n+1)xn+3a1x+\sum_{n=2}^{\infty} a_n n (n-1) x^n - 2a_2 -6a_3 x - \sum_{n=2}^{\infty} a_{n+2} (n+2) (n+1) x^n + 3a_1 x +3n=2annxn+a0x+n=2an1xn=03 \sum_{n=2}^{\infty} a_n n x^n + a_0 x + \sum_{n=2}^{\infty} a_{n-1} x^n = 0


{2a2=06a3+3a1+a0=0ann(n1)an+2(n+2)(n+1)+3ann+an1=0,n=2,3,4...\begin{cases} -2 a_2 = 0 \\ -6 a_3 + 3 a_1 + a_0 = 0 \\ a_n n (n-1) - a_{n+2} (n+2) (n+1) + 3 a_n n + a_{n-1} = 0, n=2,3,4... \end{cases}{a2=0a3=3a1+a06an+2=ann(n+2)+an1(n+2)(n+1),n=2,3,4...\begin{cases} a_2 = 0 \\ a_3 =\frac {3 a_1 + a_0} 6 \\ a_{n+2} = \frac {a_n n (n+2) + a_{n-1}} {(n+2) (n+1)}, n=2,3,4... \end{cases}

Answer:

y(x)=a0+a1x+3a1+a06x3+n=2ann(n+2)+an1(n+2)(n+1)xn+2y(x) = a_0 +a_1 x + \frac {3 a_1 + a_0} {6} x^3 + \sum_{n=2}^{\infty} \frac {a_n n (n+2) + a_{n-1}} {(n+2)(n+1)} x^{n+2}


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