Question

Differential Equation
Need help finding recursive equation
(1+x^2)-y'+y=0

Answer 1

These are my work

Answer 2

\[x ^{0}[2a_{2}-a_{1}+a_{0}]+x ^{1}[3*2a_{3}+a_{1}] +\sum_{k=2}^{\infty}[(k+2)(k+1)a_{k+2}-(k+1)a_{ak+1}+a_{k}+k(k-1)a_{K}]x ^{k}\]

Answer 3

x ^{0}[2a_{2}-a_{1}+a_{0}]+x ^{1}[3*2a_{3}+a_{1}]
+sum_{k=2}^{infty}[(k+2)(k+1)a_{k+2}-(k+1)a_{ak+1}+a_{k}+k(k-1)a_{K}]x ^{k}

Pattern
\[2a_{2}=a_{1}-a_{0}\]
\[[3*2]a_{3}=2a_{2}-a_{1}\]
\[[4*3]a_{4}=3a_{3}-a_{2}[1+(2)(1)]\]
\[[5*4]a_{5}=4a_{4}-a_{3}[1+(3)(2)]\]
\[[6*5]a_{6}=5a_{5}-a_{4}[1+(4)(3)]\]
\[[7*6]a_{7}=6a_{6}-a_{5}[1+(5)(4)]\]

Answer 4

\[(1+x^2)y''-y'+y=0\]
It looks like you're looking for a solution about \(x=0\), so
\[y=\sum_{n=0}^\infty a_nx^n\\
y'=\sum_{n=1}^\infty na_nx^{n-1}\\
y'=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\]
\[\begin{align*}(1+x^2)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^n&=0\\
\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=2}^\infty n(n-1)a_nx^n-\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^n&=0
\end{align*}\]
Let's write out some terms from the left hand side:
\[\begin{matrix}\underline{\text{power of } x}&&&\underline{\text{coefficient}}\\
\text{constant }(0)&&&2\cdot1a_2-1a_1+a_0\\
1&&&3\cdot2a_3-2a_2+a_1\\
2&&&4\cdot3a_4-3a_3+a_2\\
3&&&5\cdot4a_5-4a_4+a_3\\
\vdots&&&\vdots\\
n&&&(n+2)(n+1)a_{n+2}-(n+1)a_{n+1}+a_n\end{matrix}\]
Matching the same-power terms on the left and right sides of the equation tells you that each coefficient in the table above is set equal to zero.

Were you given any initial conditions? It would make finding a closed form of the formula for \(a_n\) much simpler.

Answer 5

Nope, no initial condition all that's given is a_2 and a_3
ah well, supposedly there is no pattern for this if there was, it require ridiculous method beyond ordinary differential equation