$$xdy = ({y^2} - 4y)dx = 0$$

$$ \Rightarrow xdy = ({y^2} - 4y)dx$$

$$ \Rightarrow {{dy} \over {({y^2} - 4x)}} = {{dx} \over x}$$

$$ \Rightarrow {{dy} \over {y(y - 4)}} = {{dx} \over x}$$

$$ \Rightarrow {1 \over 4} \times {{y - (y - 4)} \over {y(y - 4)}}dy = {{dx} \over x}$$

$$ \Rightarrow {1 \over 4}\left( {{1 \over {y - 4}} - {1 \over y}} \right)dy = {{dx} \over x}$$

Integrating both side, we get

$$ \Rightarrow {1 \over 4}\int {\left( {{1 \over {y - 4}} - {1 \over y}} \right)dy = \int {{{dx} \over x}} } $$

$$ \Rightarrow {1 \over 4}\left( {\ln \left| {{{y - 4} \over y}} \right|} \right) = \ln |x| + {1 \over 4}\ln C$$ (Cont.)

$$ \Rightarrow {1 \over 4}\ln \left| {{{y - 4} \over y}} \right| = \ln x + {1 \over 4}\ln C$$ [As $$x > 0$$, so $$|x| = x$$]

$$ \Rightarrow \ln \left| {{{y - 4} \over y}} \right| = 4\ln x + \ln C$$

$$ \Rightarrow \ln \left| {{{y - 4} \over y}} \right| = \ln ({x^4}C)$$

$$ \Rightarrow {{y - 4} \over y} = \, \pm \,C{x^4}$$

$$ \Rightarrow {{y - 4} \over y} = \lambda {x^4}$$ [Assume $$ \pm \,C = \lambda $$]

Given, $$y(1) = 2$$

$$\therefore$$ $${{2 - 4} \over 2} = \lambda {(1)^4}$$

$$ \Rightarrow \lambda = -1$$

$$\therefore$$ $${{y - 4} \over y} = - {x^4}$$

Now, $$y(\sqrt 2 ) = - {(\sqrt 2 )^4} = {{y - 4} \over y}$$

$$ \Rightarrow {{y - 4} \over y} = - 4$$

$$ \Rightarrow y - 4 = - 4y$$

$$ \Rightarrow 5y = 4$$

$$ \Rightarrow y = {4 \over 5}$$

$$\therefore$$ $$10y(\sqrt 2 ) = 10 \times {4 \over 5} = 8$$