JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 2)
If y = y(x) satisfies the differential equation
$${8\sqrt x \left( {\sqrt {9 + \sqrt x } } \right)dy = {{\left( {\sqrt {4 + \sqrt {9 + \sqrt x } } } \right)}^{ - 1}}}$$
dx, x > 0 and y(0) = $$\sqrt 7 $$, then y(256) =
$${8\sqrt x \left( {\sqrt {9 + \sqrt x } } \right)dy = {{\left( {\sqrt {4 + \sqrt {9 + \sqrt x } } } \right)}^{ - 1}}}$$
dx, x > 0 and y(0) = $$\sqrt 7 $$, then y(256) =
16
3
9
80
Explanation
$${{dy} \over {dx}} = {1 \over {8\sqrt x + \sqrt {9 + \sqrt x } \sqrt {4 + \sqrt {9 + \sqrt x } } }}$$
$$ \Rightarrow y = \sqrt {4 + \sqrt { + \sqrt x } } + c$$
Now, $$y(0) = \sqrt 7 + c$$
$$ \Rightarrow c = 0$$
$$y(256) = \sqrt {4 + \sqrt {9 + \sqrt {16} } } = \sqrt {4 + 5} = 3$$
$$ \Rightarrow y = \sqrt {4 + \sqrt { + \sqrt x } } + c$$
Now, $$y(0) = \sqrt 7 + c$$
$$ \Rightarrow c = 0$$
$$y(256) = \sqrt {4 + \sqrt {9 + \sqrt {16} } } = \sqrt {4 + 5} = 3$$
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