Subject: Applied Mathematics 4

Difficulty: Medium

Marks: 05 Marks

Let F= $ \frac{1+ y^2}{y^{12}} - - - - 1$

F does not contain x explicity, we use the formula,

$ F- Y' (\frac{SF}{SY'}) = C - - - - 2$

Diff 1 w.r.t y'

$ \frac{SF}{SY'} = \frac{y^{12}(0)-(1+y^2)(2y')}{Y'^4}$

$ \frac{SF}{SY'} = \frac{-(1+y^2)(2y')}{Y'^4}$

$ \frac{SF}{SY'} = \frac{-2(1+y^2)}{Y'^3} - - - - 3$

put equation 3 and equation 1 in equation 2

$ \begin{align} \frac{(1+y^2)}{Y'^2} - y' [\frac{-2(1+y^2)}{Y'^3}] &= c \\ \frac{(1+y^2)}{Y'^2} + \frac{2(1+y^2)}{Y'^2} &= c \\ \frac{3(1+y^2)}{Y'^2} &= c\\ \frac{Y'^2}{3(1+y^2)} &= \frac{1}{c} \\ Y'^2 &= \frac{3(1+y^2)}{c} \\ Y' &= \sqrt{\frac{3}{c}(1+y^2)} \\ \end{align}$