If $y = 3^{x}\arcsin(x)$ prove that $(1 - x^2)y'' - xy' = 0$
Our professor got this question into two examination cycles and this could be his third, YET no matter how much i try this just doesn't work?? If someone can enlighten me I'd be more than grateful.
(Sorry if the format of the question is weird, I haven't used this website before so I am unfamiliar with how it works)
A simple calculation show that $y=3^{x}\arcsin(x)$ is not a solution for the equation $(1-x^2)y^{\prime \prime}-xy^{\prime}=0$
Consider $$y^{\prime}=3^{x} \log (3) \arcsin(x)+\frac{3^x}{\sqrt{1-x^2}}$$ $$y^{\prime \prime}=\log(3)\left( 3^{x} \log (3) \arcsin(x)+\frac{3^x}{\sqrt{1-x^2}} \right)+3^{x} \log (3)\frac{1}{\sqrt{1-x^2}}+\frac{x(3^x)}{\sqrt{1-x^2}}$$ and hence the first part is $$y^{\prime \prime}(1-x^2)=3^{x}\log^2(x)\arcsin(x)(1-x^2)+3^{x}\log(3)(1-x^2)^{1/2}+3^{x}\log(3)(1-x^2)^{1/2}+\frac{x(3^x)}{\sqrt{1-x^2}}$$ and the second part $$xy^{\prime}=(3^x)x \log(3)\arcsin(x)+\frac{x(3^x)}{\sqrt{1-x^2}}$$ Notice that the first part is different from the second, and hence $y$ can`t be a solution for the equation.
For find a solution, you can solve the differential equation $$(1-x^2)y^{\prime \prime}-xy^{\prime}=0$$ for $y$ by substitution, and if you do the correct calculation you should get as solution $$y=c_1\arcsin(x)+c_2$$ where $c_1,c_2\in \mathbb{R}$.