Question: In Problems 31-40, solve the initial value problem.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\left(\frac{x}{y}\mathbf{+}\frac{y}{x}\right)\mathbf{,}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{4}$

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\left(\frac{x}{y}\mathbf{+}\frac{y}{x}\right)\mathbf{,}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{4}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Evaluate the given equation.

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\left(\frac{x}{y}\mathbf{+}\frac{y}{x}\right)\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{y}{x}\mathbf{=}\frac{x}{y}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{y}{x}\mathbf{=}{\mathbf{xy}}^{-1}\end{array}$

Since, the given equation is the Bernoulli equation with n = -1,$\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{1}{x}$ and Q ( x ) = x.

Now divide y^-1 on both sides of the equation (1).

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

Let us take u = y² and $\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{y}\frac{\mathrm{dy}}{\mathrm{dx}}$. Then,

Let $\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{2}{x}$.

Find the value of $\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\int \frac{2}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ln}\left|x\right|}\\ \mathbf{=}\frac{1}{{\mathbf{x}}^2}\end{array}$

Multiply $\frac{1}{{\mathbf{x}}^2}$in equation (3) on both sides.

$\begin{array}{c}\frac{1}{{\mathbf{x}}^2}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{-}\frac{2}{{\mathbf{x}}^3}\mathbf{u}\mathbf{=}\frac{2}{x}\\ \frac{\mathrm{du}}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^2}\mathbf{u}\right]\mathbf{=}\frac{2}{x}\end{array}$

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{\mathrm{du}}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^2}\mathbf{u}\right]\mathbf{dx}\mathbf{=}\int \frac{2}{x}\mathbf{dx}\\ \frac{1}{{\mathbf{x}}^2}\mathbf{u}\mathbf{=}\mathbf{2}\mathbf{ln}\left|x\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{u}\mathbf{=}\mathbf{2}{\mathbf{x}}^2\mathbf{ln}\left|x\right|\mathbf{+}{\mathbf{x}}^2\mathbf{C}\\ {\mathbf{y}}^2\mathbf{=}\mathbf{2}{\mathbf{x}}^2\mathbf{ln}\left|x\right|\mathbf{+}{\mathbf{x}}^2\mathbf{C}\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)\end{array}$

So, the solution is found.

Given that, y ( 1 ) = -4.

Then, x = 1 and y = -4.

Substitute the value in equation (4) to get the value of C.

Substitute the value of C in equation (2).

$\begin{array}{c}{\mathbf{y}}^2\mathbf{=}\mathbf{2}{\mathbf{x}}^2\mathbf{ln}\left|x\right|\mathbf{+}16{\mathbf{x}}^2\\ \mathbf{y}\mathbf{=}\sqrt{\mathbf{2}{\mathbf{x}}^2\mathbf{ln}\left|x\right|\mathbf{+}16{\mathbf{x}}^2}\end{array}$

So, the solution is $\mathbf{y}\mathbf{=}\sqrt{\mathbf{2}{\mathbf{x}}^2\mathbf{ln}\left|x\right|\mathbf{+}16{\mathbf{x}}^2}$