If the curve $y=y(x)$ is the solution of the differential equation $2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0$ which passes through the point $\left(1,1-\frac{4}{3} \log _{e} 2\right)$, then the value of $y(16)$ is equal to:
Correct Option: , 3
$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{2 \mathrm{x}}=\frac{\mathrm{x}^{9 / 4}}{\mathrm{x}^{5 / 4}\left(\mathrm{x}^{3 / 4}+1\right)}$
$\mathrm{IF}=\mathrm{e}^{-\int \frac{\mathrm{d} \mathrm{x}}{2 \mathrm{~d}}}=\mathrm{e}^{-\frac{1}{2} \ln \mathrm{x}}=\frac{1}{\mathrm{x}^{1 / 2}}$
$y \cdot x^{-1 / 2}=\int \frac{x^{9 / 4} \cdot x^{-1 / 2}}{x^{5 / 4}\left(x^{3 / 4}+1\right)} d x$
$\mathrm{y} \cdot \mathrm{x}^{-1 / 2}=\int \frac{\mathrm{x}^{9 / 4} \cdot \mathrm{x}^{-1 / 2}}{\mathrm{x}^{5 / 4}\left(\mathrm{x}^{3 / 4}+1\right)} \mathrm{d} \mathrm{x}$
$\int \frac{x^{1 / 2}}{\left(x^{3 / 4}+1\right)} d x$
$x=t^{4} \Rightarrow d x=4 t^{3} d t$
$\int \frac{t^{2}+4 t^{3} d t}{\left(t^{3}+1\right)}$
$4 \int \frac{t^{2}\left(t^{3}+1-1\right)}{\left(t^{3}+1\right)} d t$
$\frac{4 t^{3}}{3}-\frac{4}{3} \ln \left(t^{3}+1\right)+C$
$y x^{-1 / 2}=\frac{4 x^{3 / 4}}{3}-\frac{4}{3} \ln \left(x^{3 / 4}+1\right)+C$
$1-\frac{4}{3} \log _{e} 2=\frac{4}{3}-\frac{4}{3} \log _{e} 2+C$
$\Rightarrow C=-\frac{1}{3}$
$y=\frac{4}{3} x^{5 / 4}-\frac{4}{3} \sqrt{x} \ln \left(x^{3 / 4}+1\right)-\frac{\sqrt{x}}{3}$
$y(16)=\frac{4}{3} \times 32-\frac{4}{3} \times 4 \ln 9-\frac{4}{3}$
$=\frac{124}{3}-\frac{32}{3} \ln 3=4\left(\frac{31}{3}-\frac{8}{3} \ln 3\right)$
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