Question

((x^2) + 2)y′′ + xy' − 3y = 0

Answer 1

$(x^2+2)y''+xy'-3y=0\\ \text{Let } t=\sinh^{-1} (\frac{x}{\sqrt{2}})\\ \implies x=\sqrt{2}\sinh{(t})\\ \frac{dx}{dt}=\sqrt{2}\cosh{(t})\\ \implies \frac{dt}{dx}=\frac{1}{\sqrt{x^2+2}}\\ \implies \frac{d^2t}{dx^2}=\frac{-x}{(x^2+2)^{\frac{3}{2}}}\\ \frac{dy}{dx}=\frac{1}{\sqrt{x^2+2}}\frac{dy}{dt}\\ \frac{d^2y}{dx^2}=\frac{1}{x^2+2}\left[\frac{d^2y}{dt^2}-\frac{x}{(x^2+2)^{\frac{1}{2}}}\frac{dy}{dt}\right]\\ \text{Substituting into the original equation, we have }\\ \frac{d^2y}{dt^2}-3y=0\\ \text{Get the auxiliary equation}\\ m^2-3=0\\ \implies m_1=\sqrt{3}, m_2=-\sqrt{3}\\ y=C_1e^{\sqrt{3}t}+C_2e^{-\sqrt{3}t}\\ \text{But } t= \sinh^{-1} (\frac{x}{\sqrt{2}})\\ \implies y(x)=C_1e^{\sqrt{3}\sinh^{-1} (\frac{x}{\sqrt{2}})}+C_2e^{-\sqrt{3}\sinh^{-1} (\frac{x}{\sqrt{2}})}$