1. The problem statement, all variables and given/known data
Reason or prove whether there exist polynomials A, B and C such that the following is satisfied where [itex]y=e^{k\cdot arcsinx}[/itex]:
[itex]A\cdot y''+B\cdot y'+C\cdot y=0[/itex]
Note that this is high school level calculus so it shouldn't be something too complicated. While I said "prove", "reason" is more accurate, so no need to prove formally.
The exercise is in Spanish, section 3. a) in the attachment.
2. Relevant equations
Relevant equations would be the derivatives but it's pointless to write them all here.
3. The attempt at a solution
I tried deriving y to find y' and y'' and then I substituted:
[itex]\left\{\begin{matrix}y=e^{k\cdot arcsinx}
\\ y'=e^{k\cdot arcsinx}\cdot \frac{k}{\sqrt{1+x^2}}=y\cdot \frac{k}{\sqrt{1+x^2}}
\\ y''=e^{k\cdot arcsinx}\cdot \frac{k^2}{1+x^2}+e^{k\cdot arcsinx}\cdot \frac{-kx}{\sqrt{(1+x^2)^{3}}}=y\cdot (\frac{k^2}{1+x^2}-\frac{kx}{(1+x^2)\sqrt{1+x^2}})=y\cdot (\frac{k^2\sqrt{1+x^2}-kx}{(1+x^2)\sqrt{1+x^2}})\end{matrix}\right.[/itex]
[itex]\left\{\begin{matrix}A\cdot y''+B\cdot y'+C\cdot y=0
\\ A\cdot [y\cdot (\frac{k^2\sqrt{1+x^2}-kx}{(1+x^2)\sqrt{1+x^2}})]+B\cdot [y\cdot \frac{k}{\sqrt{1+x^2}}]+C\cdot y=0
\\ A\cdot({k^2\sqrt{1+x^2}-kx})+B\cdot{k(1+x^2)}+C\cdot (1+x^2)\sqrt{1+x^2}=0\end{matrix}\right.[/itex]
Please note that I don't know if doing all the above was the best option. I tried to simplify as much as I could but perhaps I needn't have done that (ie, the solution is somewhere else).
I don't know how to explain why there should exist polynomials such that they satisfy the condition mentioned previously, so please give me some pointers.
Thank you for reading. Have a nice day.
Reason or prove whether there exist polynomials A, B and C such that the following is satisfied where [itex]y=e^{k\cdot arcsinx}[/itex]:
[itex]A\cdot y''+B\cdot y'+C\cdot y=0[/itex]
Note that this is high school level calculus so it shouldn't be something too complicated. While I said "prove", "reason" is more accurate, so no need to prove formally.
The exercise is in Spanish, section 3. a) in the attachment.
2. Relevant equations
Relevant equations would be the derivatives but it's pointless to write them all here.
3. The attempt at a solution
I tried deriving y to find y' and y'' and then I substituted:
[itex]\left\{\begin{matrix}y=e^{k\cdot arcsinx}
\\ y'=e^{k\cdot arcsinx}\cdot \frac{k}{\sqrt{1+x^2}}=y\cdot \frac{k}{\sqrt{1+x^2}}
\\ y''=e^{k\cdot arcsinx}\cdot \frac{k^2}{1+x^2}+e^{k\cdot arcsinx}\cdot \frac{-kx}{\sqrt{(1+x^2)^{3}}}=y\cdot (\frac{k^2}{1+x^2}-\frac{kx}{(1+x^2)\sqrt{1+x^2}})=y\cdot (\frac{k^2\sqrt{1+x^2}-kx}{(1+x^2)\sqrt{1+x^2}})\end{matrix}\right.[/itex]
[itex]\left\{\begin{matrix}A\cdot y''+B\cdot y'+C\cdot y=0
\\ A\cdot [y\cdot (\frac{k^2\sqrt{1+x^2}-kx}{(1+x^2)\sqrt{1+x^2}})]+B\cdot [y\cdot \frac{k}{\sqrt{1+x^2}}]+C\cdot y=0
\\ A\cdot({k^2\sqrt{1+x^2}-kx})+B\cdot{k(1+x^2)}+C\cdot (1+x^2)\sqrt{1+x^2}=0\end{matrix}\right.[/itex]
Please note that I don't know if doing all the above was the best option. I tried to simplify as much as I could but perhaps I needn't have done that (ie, the solution is somewhere else).
I don't know how to explain why there should exist polynomials such that they satisfy the condition mentioned previously, so please give me some pointers.
Thank you for reading. Have a nice day.
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