question_answer

Show that the function \[y=Ax+\frac{B}{x}\] is a solution of the differential equation \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}-y=0.\]

Answer:

We have, \[y=Ax+\frac{B}{x}\] On differentiating w.r.t. x, we get \[\frac{dy}{dx}=A-\frac{B}{{{x}^{2}}}\Rightarrow x\frac{dy}{dx}=Ax-\frac{B}{x}\] \[\Rightarrow \]   \[y+x\frac{dy}{dx}=Ax+\frac{B}{x}+Ax-\frac{B}{x}=2Ax\] ?(i) Again differentiating, we get \[\frac{dy}{dx}+x\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}=2A\] \[\Rightarrow \]   \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x\frac{dy}{dx}=2Ax\] \[\Rightarrow \]   \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x\frac{dy}{dx}=y+x\frac{dy}{dx}\]  [from Eq. (i)] \[\Rightarrow \]   \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}-y=0\]