In problem 7-16, solve the equation.$\mathbf{x}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{1}{{\mathbf{y}}^3}$

A first-order ordinary differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions $\mathbf{g}\left(x\right)$that is a function of x alone and $\mathbf{h}\left(y\right)$that is a function of y alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{h}\left(y\right)$. By separating the variables, the equation follows $\frac{\mathrm{dy}}{\mathbf{h}\left(y\right)}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{dx}$. Then, on direct integration of both sides, the solution of the differential equation is determined.

The given equation is

$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{{\mathrm{y}}^3}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

After separating the variables, equation (1) can be written as

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

Integrate both sides of equation (2). It results,

$\begin{array}{c}\int {\mathrm{y}}^3\mathrm{dy}=\int \frac{dx}{x}\\ \frac{{\mathrm{y}}^4}{4}=\ln \mathrm{x}+\mathrm{k}\left[\mathrm{k}=\mathrm{IntegratingConstant}\right]\\ {\mathrm{y}}^4=4\ln \mathrm{x}+4\mathrm{k}\\ {\mathrm{y}}^4=\ln {\mathrm{x}}^4+\mathrm{C}\left[\mathrm{C}=4\mathrm{k}=\mathrm{constant},\mathrm{n}\ln \mathrm{x}=\ln {\mathrm{x}}^{\mathrm{n}}\right]\end{array}$

$\begin{array}{c}\mathrm{y}=\pm {\left[\ln {\mathrm{x}}^4+\mathrm{C}\right]}^{\frac{1}{4}}\\ \mathrm{y}=\pm \sqrt[4]{\ln \left({\mathrm{x}}^4\right)+\mathrm{C}}\end{array}$

Therefore, the solution of the given equation is $\mathbf{y}\mathbf{=}\mathbf{\pm }\sqrt[4]{\mathbf{ln}\left({\mathbf{x}}^4\right)\mathbf{+}\mathbf{C}}$