JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 6)

If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2$$, then :
$${x^2}y'' + xy' - 25y = 0$$
$${x^2}y'' - xy' - 25y = 0$$
$${x^2}y'' - xy' + 25y = 0$$
$${x^2}y'' + xy' + 25y = 0$$

Explanation

$${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$$

Differentiating on both side

$$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$$

$${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} $$

Square on both side

$${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$$

Diff on both side

$$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$$

$$xy' + y''{x^2} + 25y = 0$$

Comments (0)

Advertisement