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Given that for an equation,

(18) $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{P}\left(x\right){\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{Q}\left(x\right)\mathbf{y}\mathbf{+}\mathbf{R}\left(x\right)$

one solution u(x)鈥攐f (18) is known

Therefore,

$\begin{array}{c}\frac{\mathbf{du}}{\mathbf{dx}}\mathbf{=}{\mathbf{Pu}}^{\mathbf{2}}\mathbf{+}\mathbf{Qu}\mathbf{+}\mathbf{R}\\ \frac{\mathbf{du}}{\mathbf{dx}}\mathbf{-}{\mathbf{Pu}}^{\mathbf{2}}\mathbf{-}\mathbf{Qu}\mathbf{-}\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\end{array}$

Now,$\mathbf{y}\mathbf{=}\mathbf{u}\mathbf{+}\frac{1}{v}$

Differentiating both sides with respect to x,

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{du}}{\mathbf{dx}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\frac{\mathbf{dv}}{\mathbf{dx}}$

Substitute $\mathbf{y}\mathbf{=}\mathbf{u}\mathbf{+}\frac{1}{v}$in equation (18),

$\begin{array}{c}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{P}{\left(u+\frac{1}{v}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{Q}\left(u+\frac{1}{v}\right)\mathbf{+}\mathbf{R}\\ \mathbf{du}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\mathbf{P}{\left(u+\frac{1}{v}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{Q}\left(u+\frac{1}{v}\right)\mathbf{+}\mathbf{R}\end{array}$

Now using equation (1),

$\begin{array}{c}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{P}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}\mathbf{Pu}}{\mathbf{v}}\mathbf{+}\frac{\mathbf{Q}}{\mathbf{v}}\\ \mathbf{-}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{-}\mathbf{P}\frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}\mathbf{Pu}}{\mathbf{v}}\mathbf{-}\frac{\mathbf{Q}}{\mathbf{v}}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\mathbf{P}\mathbf{-}\mathbf{2}\mathbf{Puv}\mathbf{-}\mathbf{Qv}\\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\left(2\mathrm{Pu}+Q\right)\mathbf{v}\mathbf{-}\mathbf{P}\\ {\mathbf{v}}^{\mathbf{\text{'}}}\mathbf{+}\left(2\mathrm{Pu}+Q\right)\mathbf{v}\mathbf{=}\mathbf{-}\mathbf{P}\end{array}$

Hence, the substitution $\mathbf{y}\mathbf{=}\mathbf{u}\mathbf{+}\frac{1}{v}$reduces (18) to a linear equation in v i.e., ${\mathit{v}}^{\mathbf{\text{'}}}\mathbf{+}\left(2Pu+Q\right)\mathit{v}\mathbf{=}\mathbf{-}\mathit{P}$

The given equation is,

$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^3{\left(\mathrm{y}-\mathrm{x}\right)}^2+\frac{\mathrm{y}}{\mathrm{x}}$

Simplifying this equation,

$\begin{array}{c}\frac{dy}{dx}=x^3{\left(y-x\right)}^2+\frac{y}{x}\\ =x^3\left(y^2+x^2-2xy\right)+\frac{y}{x}\\ =x^3{y}^2+x^5-2x^{4}y+\frac{y}{x}\\ =x^3{y}^2+\left(\frac{1}{x}-2x^4\right)y+x^5\end{array}$

Comparing this equation with $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{P}\left(\mathrm{x}\right){\mathrm{y}}^2+\mathrm{Q}\left(\mathrm{x}\right)\mathrm{y}+\mathrm{R}\left(\mathrm{x}\right)$, one will get,

$\mathrm{P}={\mathrm{x}}^3,\mathrm{Q}=\frac{1}{\mathrm{x}}-2{\mathrm{x}}^4,\mathrm{R}={\mathrm{x}}^5$

Substituting these values of P, Q and R in the result of part (a) in step 2,

$\frac{\mathrm{dv}}{\mathrm{dx}}+\left(2{\mathrm{x}}^3\mathrm{u}+\frac{1}{\mathrm{x}}-2{\mathrm{x}}^4\right)\mathrm{v}=-{\mathrm{x}}^3.....\left(2\right)$

Now as u(x) = x is a solution of equation (2), therefore (2) becomes,

$\begin{array}{c}\frac{dv}{dx}+\left(2x^4+\frac{1}{x}-2x^4\right)v=-x^3\\ \frac{dv}{dx}+\left(\frac{1}{x}\right)v=-x^3\mathrm{.....}\left(3\right)\end{array}$

Now equation (3) is of form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

Therefore, integrating factor =$e^{鈭�\frac{1}{x}dx}=e^{lnx}=x$

Multiplying equation (3) by x,

$\begin{array}{c}x\frac{dv}{dx}+v=-x^4\\ \frac{d}{dx}\left(v鈰�x\right)=-x^4\end{array}$

Integrating both sides,

$vx=-\frac{x^5}{5}+C\mathrm{......}\left(4\right)$

Where, C is the constant of integration.

Now using the substitution given in the question $\mathbf{y}\mathbf{=}\mathbf{u}\mathbf{+}\frac{1}{v}$in part (a), one will get the value of v,

$v=\frac{1}{y-x}$

Substitute this value of v in equation (4),

$\begin{array}{c}\frac{x}{y-x}=-\frac{x^5}{5}+C_1\\ \frac{x}{y-x}=-\frac{x^5+5C_1}{5}\\ y=x+\frac{5x}{C-x^5}\end{array}$

Where, C = 5C is an arbitrary constant.

Hence, the solution of the equation $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}{\left(y-x\right)}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\mathrm{is}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{+}\frac{\mathbf{5}\mathbf{x}}{\mathbf{C}\mathbf{-}{\mathbf{x}}^{\mathbf{5}}}$