If the curve y = y(x) is the solution of the differential equation

$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$, x > 0 which

passes through the point $\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$, then the value of y(16) is equal to :

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