If the curve $y=y\left(x\right)$ is the solution of the differential equation $2\left(x^2+x^{5/4}\right)dy-y\left(x+x^{1/4}\right)dx=2x^{9/4}dx, x>0$ which passes through the point $\left(1, 1-\frac{4}{3}{\log }_{\mathrm{e}}2\right),$ then the value of $y\left(16\right)$ is equal to 

Given differential equation is $2\left(x^2+x^{5/4}\right)dy-y\left(x+x^{1/4}\right)dx=2x^{9/4}dx$

$\Rightarrow 2\left(x^2+x^{5/4}\right)\frac{dy}{dx}-y\left(x+x^{1/4}\right)=2x^{9/4}$

$\Rightarrow \frac{dy}{dx}-\frac{y\left(x+x^{1/4}\right)}{2\left(x^2+x^{5/4}\right)}=\frac{2x^{9/4}}{2\left(x^2+x^{5/4}\right)}$

$\Rightarrow \frac{dy}{dx}-\frac{y\left(x+x^{1/4}\right)}{2x\left(x+x^{1/4}\right)}=\frac{2x^{9/4}}{2x^{5/4}\left(x^{3/4}+1\right)}$

$\Rightarrow \frac{dy}{dx}-\frac{y}{2x}=\frac{x}{\left(x^{3/4}+1\right)}$

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q,$ where $P=-\frac{1}{2x}$ and $Q=\frac{x}{x^{3/4}+1}$

Now, we have integrating factor $I.F.=e^{\int Pdx}$

$=e^{-\int \frac{dx}{2x}}=e^{-\frac{1}{2}{\log }_{e}x}$

$=e^{{\log }_e{x}^{-\frac{1}{2}}}=x^{-\frac{1}{2}}$

Now, the solution of the linear differential equation is $y\times \left(I.F.\right)=\int Q\times \left(I.F.\right)dx+C$

$\Rightarrow y·x^{-1/2}=\int \frac{x·x^{-1/2}}{\left(x^{3/4}+1\right)}dx+C$

$\Rightarrow y·x^{-1/2} =\int \frac{x^{1/2}}{\left(x^{3/4}+1\right)}dx+C$

Put $x=t^4\Rightarrow dx=4t^{3}dt$

$\Rightarrow y·x^{-1/2} =\int \frac{t^2·4t^{3}dt}{\left(t^3+1\right)}$

$\Rightarrow y·x^{-1/2} =4\int \frac{t^2\left(t^3+1-1\right)}{\left(t^3+1\right)}dt$

$\Rightarrow y·x^{-1/2} =4\int t^{2}dt-4\int \frac{t^2}{t^3+1}dt+C$

Let, $t^3+1=u \Rightarrow 3t^{2}dt=du$

$\Rightarrow y·x^{-1/2} =4\int t^{2}dt-\frac{4}{3}\int \frac{du}{u}+C$

$\Rightarrow y·x^{-1/2} =\frac{4t^3}{3}-\frac{4}{3}{\log }_e\left|u\right|+C$

$\Rightarrow y·x^{-1/2} =\frac{4t^3}{3}-\frac{4}{3}{\log }_e\left(t^3+1\right)+C$

$\Rightarrow y·x^{-1/2}=\frac{4x^{3/4}}{3}-\frac{4}{3}{\log }_e\left(x^{3/4}+1\right)+C$

Now, the curve passes through $\left(1, 1-\frac{4}{3}{\log }_{\mathrm{e}}2\right)$

$\Rightarrow 1-\frac{4}{3}{\log }_{\mathrm{e}}2=\frac{4}{3}-\frac{4}{3}{\log }_{\mathrm{e}}2+C$

$\Rightarrow C=-\frac{1}{3}$

$\Rightarrow y·x^{-1/2}=\frac{4}{3}{x}^{3/4}-\frac{4}{3}{\log }_e\left(x^{3/4}+1\right)-\frac{1}{3}$

$\Rightarrow y=\frac{4}{3}{x}^{5/4}-\frac{4}{3}\sqrt{x}{\log }_e\left(x^{3/4}+1\right)-\frac{\sqrt{x}}{3}$

Hence, at $x=16,$ we get 

$y\left(16\right)=\frac{4}{3}\times 32-\frac{4}{3}\times 4\times {\log }_{e}9-\frac{4}{3}$

$\Rightarrow y\left(16\right)=\frac{124}{3}-\frac{32}{3}{\log }_{e}3$

$\Rightarrow y\left(16\right)=4\left(\frac{31}{3}-\frac{8}{3}{\log }_{e}3\right).$