Solve the following differential equations by method (a) or (b) above.

$x{y}^{\text{'}\text{'}}=y^{\text{'}}+y^{\text{'}3}$

The differential equation is $x{y}^{\text{'}\text{'}}=y^{\text{'}}+y^{\text{'}3}$.

A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of the function define the rate of change of a function at a point.

The given equation is lacking by dependent variable $y$, thus it can be significantly simplified by using the substitution$y^{\text{'}}\equiv p$, also imply $y^{\text{'}\text{'}}=p^{\text{'}}$. Inserting this transformation into the equation,

$x{p}^{\text{'}}=p+p^3$

The previous equation is separable and factorize its RHS into $p\left(1+p^2\right)$:

$x\frac{dp}{dx}=p\left(1+p^2\right)=\frac{dp}{p\left(1+p^2\right)}=\frac{dx}{x}$ …….. (1)

The integral on the RHS is table integral. Now start by factoring out $p^2$from the parenthesis in the denominator:

$\int \frac{dp}{p\left(1+p^2\right)}=\int \frac{dp}{p^3\left(1+\frac{1}{p^2}\right)}$ …….. (2)

In the above equation, the derivative of the expression in the parentheses would yield up to the constant factor, a factor outside the parentheses. Now the result after substitution as:

$$\begin{gathered} u=1+\frac{1}{p^2} \\ \Rightarrow du=-2\frac{dp}{p^3} \\ \Rightarrow \frac{dp}{p^3}=-\frac{du}{2} \end{gathered}$$

Combine equation (1) and (2),

$-\frac{du}{u}=2\frac{dx}{x}$

This form of the equation is easily integral by using the table integral $\int \frac{dx}{x}=\ln \left|x\right|+C$ and identity $a\ln x=\ln x^a$

$-\ln u=\ln x^2+A$

Now using the identity $\ln x+\ln y=\ln xy$along with the definition $u=1+\frac{1}{p^2}$

$\ln \frac{1}{1+\frac{1}{v^2}}=\ln \left(A{x}^2\right)$with$A\equiv e^A$

The previous equation is

$$\begin{gathered} \frac{1}{p^2}=\frac{1}{A{x}^2}-1 \\ \Rightarrow p=\frac{1}{\sqrt{\frac{1}{A{x}^2}-1}} \end{gathered}$$

Solve for$y$ remembering that$p=\frac{dy}{dx}$ :

$\frac{dy}{dx}=\frac{1}{\sqrt{\frac{1}{A{x}^2}-1}}\Rightarrow dy=\sqrt{A}\frac{xdx}{\sqrt{1-A{x}^2}}$

Multiply the RHS of the previous equation with$1=\frac{x\sqrt{A}}{x\sqrt{A}}$ .

Noting that the derivative of expression inside the square root yield the denominator of said RHS up to a constant factor, make the variable substitution $u=1-A{x}^2\Rightarrow du=-2Axdx$and insert it into the differential equation obtained:

$dy=-\frac{\sqrt{A}}{2A}\frac{du}{\sqrt{u}}\Rightarrow y=-\frac{\sqrt{A}}{2A}\left(2\sqrt{u}\right)+B=-\frac{1}{\sqrt{A}}\sqrt{u}+B$

By virtue of the table integral $\int x^n=\frac{1}{n+1}{x}^{n+1}+$const and with integration constant labelled by $B$. Rewriting$u$ in terms of$x$ and relabel$B\sqrt{A}\to B$ to obtain:

$y\sqrt{A}=-\sqrt{1-A{x}^2}+B\sqrt{A}\Rightarrow y=-\sqrt{\frac{1}{A}-x^2}+B$

At the end, relabel $\frac{1}{A}\to A$for further convenience

$y=-\sqrt{A-x^2}+B$

Thus, the solution of the differential equation is$y=-\sqrt{A-x^2}+B$ .