Official solution
\[{{dy} \over {dx}} - {y \over {2x}} = {{{x^{9/4}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}\]
\[IF = {e^{ - \int {{{dx} \over {2x}}} }} = {e^{ - {1 \over 2}\ln x}} = {1 \over {{x^{1/2}}}}\]
\[y.{x^{ - 1/2}} = \int {{{{x^{9/4}}.{x^{ - 1/2}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}} dx\]
= \[\int {{{{x^{1/2}}} \over {({x^{3/4}} + 1)}}} dx\]
Let, \[x = {t^4} \Rightarrow dx = 4{t^3}dt\]
= \[\int {{{{t^2}.4{t^3}dt} \over {({t^3} + 1)}}} \]
= \[4\int {{{{t^2}({t^3} + 1 - 1)} \over {({t^3} + 1)}}} d\]
= \[4\int {{t^2}dt - 4\int {{{{t^2}} \over {{t^3} + 1}}dt} } \]
= \[{{4{t^3}} \over 3} - {4 \over 3}\ln ({t^3} + 1) + C\]
\[y{x^{ - 1/2}} = {{4{x^{3/4}}} \over 3} - {4 \over 3}\ln ({x^{3/4}} + 1) + C\]
It passes through the point \[\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)\]
\[ \therefore \] \[1 - {4 \over 3}{\log _e}2 = {4 \over 3} - {4 \over 3}{\log _e}2 + C\]
\[ \Rightarrow C = - {1 \over 3}\]
\[y = {4 \over 3}{x^{5/4}} - {4 \over 3}\sqrt x \ln ({x^{3/4}} + 1) - {{\sqrt x } \over 3}\]
\[y(16) = {4 \over 3} \times 32 - {4 \over 3} \times 4\ln 9 - {4 \over 3}\]
\[ = {{124} \over 3} - {{32} \over 3}\ln 3 = 4\left( {{{31} \over 3} - {8 \over 3}\ln 3} \right)\]