Question

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

• A. $(\frac{31}{3}-\frac{8}{3}{\log }_{e}3)$
• B. $4(\frac{31}{3}+\frac{8}{3}{\log }_{e}3)$
• C. $(\frac{31}{3}+\frac{8}{3}{\log }_{e}3)$
• D. $4(\frac{31}{3}-\frac{8}{3}{\log }_{e}3)$

Answer 1

The correct option is D $4(\frac{31}{3}-\frac{8}{3}{\log }_{e}3)$

$2(x^2+x^{\frac{5}{4}})dy-y(x+x^{\frac{1}{4}})dx=2x^{\frac{9}{4}}dx,x>0$
$\Rightarrow \frac{dy}{dx}-\frac{y}{2x}=\frac{x^{\frac{9}{4}}}{x^{\frac{5}{4}}(x^{\frac{3}{4}}+1)}$
$I.F=e^{-\int \frac{dx}{2x}}=e^{-\frac{1}{2}\ln x}=\frac{1}{x^{\frac{1}{2}}}$
$y\cdot x^{-\frac{1}{2}}=\int \frac{x^{\frac{9}{4}}\cdot x^{-\frac{1}{2}}}{x^{\frac{5}{4}}(x^{\frac{3}{4}}+1)}dx$
$=\int \frac{x^{\frac{1}{2}}}{(x^{\frac{3}{4}}+1)}dx\cdots \left(i\right)$
$I=\int \frac{x^{\frac{1}{2}}}{(x^{\frac{3}{4}}+1)}dx$
Substituting $x=t^4\Rightarrow dx=4t^{3}dt$
$=\int \frac{t^2\cdot 4t^{3}dt}{(t^3+1)}$
$=4\int \frac{t^2(t^3+1-1)}{(t^3+1)}dt$
$=4\int t^{2}dt-4\int \frac{t^2}{(t^3+1)}dt$
$=\frac{4t^3}{3}-\frac{4}{3}\ln (t^3+1)+C$
From $\left(i\right),$
$y{x}^{-\frac{1}{2}}=\frac{4x^{\frac{3}{4}}}{3}-\frac{4}{3}\ln \left(x^{\frac{3}{4}+1}\right)+C$
As curve passes through $(1,1-\frac{1}{4}{\log }_{e}2)$
$\Rightarrow 1-\frac{4}{3}{\log }_{e}2=\frac{4}{3}-\frac{4}{3}{\log }_{e}2+C$
$\Rightarrow C=-\frac{1}{3}$

$\therefore$ Curve equation is$y=\frac{4}{3}{x}^{\frac{5}{4}}-\frac{4}{3}\sqrt{x}\ln (x^{\frac{3}{4}}+1)-\frac{\sqrt{x}}{3}$
Now ,$y\left(16\right)=\frac{4}{3}\times 32-\frac{4}{3}\times 4\ln 9-\frac{4}{3}$
$=\frac{124}{3}-\frac{32}{3}\ln 3=4(\frac{31}{3}-\frac{8}{3}\ln 3)$