Question

Y=e^tan^-1x prove that (1+x^2)y2+(2x-1)y1=0 in differential

Answer 1

Given,

\$y = e^{\tan^{-1}x}  \\ \\ \text{Diff. both sides  with respect  to x}  \\ \\  y_1  = \frac{d(e^{\tan^{-1}x)}   }{dx} \\ \\   y_1  =  e^{\tan^{-1}x} \  \frac{d \tan^{-1}x}{dx} \\ \\ y_1=   e^{\tan^{-1}x} .\frac{1}{1 + x^{2}} \\ \\   \text{Diff. both sides  with respect  to x} \\ \\ y_2 = \frac{(1+x^{2})  e^{\tan^{-1}x} - e^{\tan^{-1}x}(2x + 0) }{(1  + x^{2})^{2}}   \\ \\ (1+x^{2})}y_2 = \frac{e^{\tan^{-1}x}-2x}{1 +x^{2}} \\ \\ L.H.S \to  \\  (1+x^{2})}y_2  + (2x-1)y_1 \\ \\   \frac{(1+x^{2})e^{\tan^{-1}x}-2x.e^{\tan^{-1}x}}{1 +x^{2}}  \ + (2x-1)\frac{e^{\tan^{-1}x}}{1 +x^{2}} \\ \\ =   \frac{x^{2}.e^{\tan^{-1}x}}{ 1 + x^{2}}   \\ \\ \textbf{Note;- Check Question}$