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

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

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

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

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

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

This is a linear differential equation of the type $\dfrac{dy}{dx} + Py = Q,$ where $P = - \dfrac{1}{2x} and$Q = \dfrac{x}{{x}^{3 / 4} + 1}$</p><p>Now, we have integrating factor$I . F . = {e}^{∫ Pdx}$</p><p>$= {e}^{- ∫ \dfrac{dx}{2x}} = {e}^{- \dfrac{1}{2}\log_{e}x}$</p><p>$= {e}^{\log_{e}{x}^{- \dfrac{1}{2}}} = {x}^{- \dfrac{1}{2}}$</p><p>Now, the solution of the linear differential equation is$y \times \left(I . F .\right) = ∫ Q \times \left(I . F .\right)dx + C$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = ∫ \dfrac{x \cdot {x}^{- 1 / 2}}{\left({x}^{3 / 4} + 1\right)}dx + C$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = ∫ \dfrac{{x}^{1 / 2}}{\left({x}^{3 / 4} + 1\right)}dx + C$</p><p>Put$x = {t}^{4} ⇒ dx = 4{t}^{3}dt$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = ∫ \dfrac{{t}^{2} \cdot 4{t}^{3}dt}{\left({t}^{3} + 1\right)}$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = 4 ∫ \dfrac{{t}^{2}\left({t}^{3} + 1 - 1\right)}{\left({t}^{3} + 1\right)}dt$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = 4 ∫ {t}^{2}dt - 4 ∫ \dfrac{{t}^{2}}{{t}^{3} + 1}dt + C$</p><p>Let,${t}^{3} + 1 = u ⇒ 3{t}^{2}dt = du$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = 4 ∫ {t}^{2}dt - \dfrac{4}{3} ∫ \dfrac{du}{u} + C$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = \dfrac{4{t}^{3}}{3} - \dfrac{4}{3}\log_{e}\left|u\right| + C$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = \dfrac{4{t}^{3}}{3} - \dfrac{4}{3}\log_{e}\left({t}^{3} + 1\right) + C$</p><p>$⇒ y \cdot {x}^{- 1 / 2} = \dfrac{4{x}^{3 / 4}}{3} - \dfrac{4}{3}\log_{e}\left({x}^{3 / 4} + 1\right) + C$</p><p>Now, the curve passes through \left(1, 1 - \dfrac{4}{3}\log_{e}2\right)$

$⇒ 1 - \dfrac{4}{3}\log_{e}2 = \dfrac{4}{3} - \dfrac{4}{3}\log_{e}2 + C$

$⇒ C = - \dfrac{1}{3}$

$⇒ y \cdot {x}^{- 1 / 2} = \dfrac{4}{3}{x}^{3 / 4} - \dfrac{4}{3}\log_{e}\left({x}^{3 / 4} + 1\right) - \dfrac{1}{3}$

$⇒ y = \dfrac{4}{3}{x}^{5 / 4} - \dfrac{4}{3}\sqrt{x}\log_{e}\left({x}^{3 / 4} + 1\right) - \dfrac{\sqrt{x}}{3}$

Hence, at $x = 16,$ we get

$y ( 16 ) = \dfrac{4}{3} \times 32 - \dfrac{4}{3} \times 4 \times \log_{e}9 - \dfrac{4}{3}$

$⇒ y\left(16\right) = \dfrac{124}{3} - \dfrac{32}{3}\log_{e}3$

$⇒ y\left(16\right) = 4\left(\dfrac{31}{3} - \dfrac{8}{3}\log_{e}3\right) .</p>$