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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Question: In Problems 31-40, solve the initial value problem.$\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}\left(\frac{x}{y}\mathbf{+}\frac{y}{x}\right){\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}\left(1\right){\mathbf{=}}{\mathbf{-}}{\mathbf{4}}$

The solution of the given equation is $\mathbf{y}\mathbf{=}\sqrt{\mathbf{2}{\mathbf{x}}^{2}\mathbf{ln}\left|x\right|\mathbf{+}16{\mathbf{x}}^{2}}$.

See the step by step solution

## Step 1: Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\left(\frac{x}{y}\mathbf{+}\frac{y}{x}\right)\mathbf{,}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{4}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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,

$\begin{array}{c}\frac{1}{2}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{-}\frac{u}{x}\mathbf{=}\mathbf{x}\\ \frac{\mathrm{du}}{\mathrm{dx}}\mathbf{-}\frac{2}{x}\mathbf{u}\mathbf{=}\mathbf{2}\mathbf{x}\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

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.

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

## Step 2: integration method

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.

## Step 3: Find the initial value

Given that, y ( 1 ) = -4.

Then, x = 1 and y = -4.

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

$\begin{array}{c}{\mathbf{y}}^{2}\mathbf{=}\mathbf{2}{\mathbf{x}}^{2}\mathbf{ln}\left|x\right|\mathbf{+}{\mathbf{x}}^{2}\mathbf{C}\\ {\left(-4\right)}^{2}\mathbf{=}\mathbf{2}\left(1\right)\mathbf{ln}\left|1\right|\mathbf{+}\left(1\right)\mathbf{C}\\ 16=C\end{array}$

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