Question: In problems 33-40 Solve the equation given in Problem 3.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{y}{x}\mathbf{=}{\mathbf{x}}^3{\mathbf{y}}^2$

A first-order equation that can be written in the form,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right){\mathbf{y}}^n$,

where P(x) and Q(x) are continuous on an interval (a, b) and n is a real number, is called a Bernoulli equation.

The given equation is,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{y}{x}\mathbf{=}{\mathbf{x}}^3{\mathbf{y}}^2$

Both sides divide by y² in the above equation,

${\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{{\mathbf{y}}^{-1}}{x}\mathbf{=}{\mathbf{x}}^3\mathrm{......}\left(1\right)$

Let,

$\begin{array}{c}\mathbf{t}\mathbf{=}{\mathbf{y}}^{-1}\\ \frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}\end{array}$

Substitute the $\mathbf{t}\mathbf{=}{\mathbf{y}}^{-1}\text{\hspace{0.33em}and\hspace{0.33em}}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}$in the equation (1),

$\begin{array}{c}\mathbf{-}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{+}\frac{t}{x}\mathbf{=}{\mathbf{x}}^3\\ \frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\frac{t}{x}\mathbf{=}\mathbf{-}{\mathbf{x}}^3\mathrm{......}\left(2\right)\end{array}$

The standard form of the linear equation is,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right)$

One has,

$\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{1}{x}$

Now find the integrating factor,

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\int \frac{1}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{log}\left(x\right)}\\ \mathbf{=}\frac{1}{x}\end{array}$

Both sides Multiply by $\mu \left(x\right)\mathbf{=}\frac{1}{x}$in the equation (2),

$\frac{1}{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\left(\frac{1}{x}\right)\frac{t}{x}\mathbf{=}\mathbf{-}\left(\frac{1}{x}\right){\mathbf{x}}^3$

Simplify the above equation,

$\begin{array}{c}\frac{1}{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\frac{t}{{\mathbf{x}}^2}\mathbf{=}\mathbf{-}{\mathbf{x}}^2\\ \frac{d}{\mathrm{dx}}\left(\frac{t}{x}\right)\mathbf{=}\mathbf{-}{\mathbf{x}}^2\end{array}$

Integrate both sides and solve for the solution.

$\frac{t}{x}\mathbf{=}\mathbf{-}\int {\mathbf{x}}^2\mathbf{dx}\mathbf{+}\mathbf{c}$

Substitute t = y^-1and simplify the above equation,

$\begin{array}{c}\frac{{\mathbf{y}}^{-1}}{x}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^3}{3}\mathbf{+}\mathbf{c}\\ \frac{1}{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^4}{3}\mathbf{+}\mathbf{cx}\\ \frac{1}{y}\mathbf{=}\frac{{\mathbf{x}}^4+\mathbf{3}\mathbf{cx}}{-3}\end{array}$

Where, C = -3c

$\mathbf{y}\mathbf{=}\frac{-3}{{\mathbf{x}}^4\mathbf{+}\mathbf{Cx}}$

Hence, the solve the given equation is $\mathbf{y}\mathbf{=}\frac{-3}{{\mathbf{x}}^4\mathbf{+}\mathbf{Cx}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{y}\mathbf{\equiv }\mathbf{0}$