Question

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

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

Answer 1

The correct option is D

$4\left(\frac{31}{3}-\frac{8}{3}{\log }_{e}3\right)$

Explanation for the correct option.

Step 1: Find the integrating factor

Given a differential equation, $2\left(x^2+x^{\frac{5}{4}}\right)dy-y\left(x+x^{\frac{1}{4}}\right)dx=2x^{\frac{9}{4}}dx,x>0$.

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

Then, the integrating factor is given by:

$\begin{array}{rcl}\therefore \mathrm{IF}& =& {\mathrm{e}}^{-\int \frac{\mathrm{dx}}{2\mathrm{d}}}\\ & =& {\mathrm{e}}^{-\frac{1}{2}\ln x}\\ & =& \frac{1}{{\mathrm{x}}^{1/2}}\end{array}$

Step 2: Find the solution

$\begin{array}{rcl}\therefore y·x^{\frac{-1}{2}}& =& \int \frac{x^{{\displaystyle \frac{9}{4}}}·x^{-{\displaystyle \frac{1}{2}}}}{x^{{\displaystyle \frac{5}{4}}}\left(x^{{\displaystyle \frac{3}{4}}}+1\right)}dx\\ & =& \int \frac{x^{1/2}}{\left(x^{3/4}+1\right)}dx\end{array}$

Let, $x=t^4\Rightarrow dx=4t^{3}dt$

$\begin{array}{rcl}\mathbf{\therefore }\int \frac{t^2-4t^{3}dt}{\left(t^3+1\right)}& =& 4\int \frac{t^2\left(t^3+1-1\right)}{\left(t^3+1\right)}dt\\ & =& 4\int {\mathrm{t}}^2\mathrm{dt}-4\int \frac{{\mathrm{t}}^2}{{\mathrm{t}}^3+1}\mathrm{dt}\\ & =& \frac{4{\mathrm{t}}^3}{3}-\frac{4}{3}\ln \left({\mathrm{t}}^3+1\right)+\mathrm{C}\end{array}$

Back substituting values we get:

${\mathrm{yx}}^{-\frac{1}{2}}=\frac{4{\mathrm{x}}^{{\displaystyle \frac{3}{4}}}}{3}-\frac{4}{3}\ln \left({\mathrm{x}}^{\frac{3}{4}}+1\right)+\mathrm{C}....\left(1\right)$

Step 3: Find $y\left(16\right)$

As the curve passes through the point $\left(1,1-\frac{4}{3}{\log }_{e}2\right)$ so it satisfies equation (1):

$\begin{array}{rcl}1-\frac{4}{3}{\log }_{\mathrm{e}}2& =& \frac{4}{3}-\frac{4}{3}{\log }_{\mathrm{e}}2+\mathrm{C}\\ \mathrm{C}& =& -\frac{1}{3}\end{array}$

$\begin{array}{rcl}\therefore y& =& \frac{4}{3}{x}^{\frac{5}{4}}-\frac{4}{3}\sqrt{x}\ln \left(x^{\frac{3}{4}}+1\right)-\frac{\sqrt{x}}{3}\\ y\left(16\right)& =& \frac{4}{3}\times 32-\frac{4}{3}\times 4\ln 9-\frac{4}{3}\\ & =& \frac{124}{3}-\frac{32}{3}\ln 3\\ & =& 4\left(\frac{31}{3}-\frac{8}{3}\ln 3\right)\end{array}$

Hence, option D is correct.