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Found in: Page 76

### 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

# Use the method discussed under “Homogeneous Equations” to solve problems 9-16. $\frac{\mathbf{dx}}{\mathbf{dt}}{\mathbf{=}}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{t}\sqrt{{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}{\mathbf{tx}}$

Homogeneous equation for the given equation is $\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{t}}^{\mathbf{2}}}}{\mathbf{=}}{\mathbf{ln}}\mathbf{|}\mathbf{t}\mathbf{|}{\mathbf{+}}{\mathbf{C}}$.

See the step by step solution

## Step 1: General form of Homogeneous equation

If the right-hand side of the equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ can be expressed as a function of the ratio $\frac{\mathrm{y}}{\mathrm{x}}$ alone, then we say the equation is homogeneous.

## Step 2: Evaluate the given equation

Given, $\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{{\mathrm{x}}^{2}+\mathrm{t}\sqrt{{\mathrm{t}}^{2}+{\mathrm{x}}^{2}}}{\mathrm{tx}}$ .

Evaluate it by dividing ${\mathrm{t}}^{2}$ by both numerator and denominator.

role="math" localid="1655106866985"

## Step 3: Substitution method

Let us take $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{t}}$.

Then $\mathrm{y}=\mathrm{vt}$.

By Differentiating,

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}+\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{v}}^{2}+\sqrt{1+{\mathrm{v}}^{2}}}{\mathrm{v}}=\mathrm{v}+\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}}$

$\mathrm{v}+\frac{\sqrt{1+{\mathrm{v}}^{2}}}{\mathrm{v}}=\mathrm{v}+\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}}\phantom{\rule{0ex}{0ex}}\frac{\sqrt{1+{\mathrm{v}}^{2}}}{\mathrm{v}}=\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}}\phantom{\rule{0ex}{0ex}}\frac{\sqrt{1+{\mathrm{v}}^{2}}}{\mathrm{v}}\frac{1}{\mathrm{dv}}=\mathrm{t}\frac{1}{\mathrm{dt}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^{2}}}\mathrm{dv}=\frac{1}{\mathrm{t}}\mathrm{dt}$

## Step 4: Integrate the equation

Now, integrate on both sides.

$\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^{2}}}\mathrm{dv}=\int \frac{1}{\mathrm{t}}\mathrm{dt}\phantom{\rule{0ex}{0ex}}\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^{2}}}\mathrm{dv}=\mathrm{ln}\left|\mathrm{t}\right|+\mathrm{C}$

Integrate $\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^{2}}}\mathrm{dv}$ separately.

Let us take $\mathrm{w}=1+{\mathrm{v}}^{2}$.

Then, role="math" localid="1655108898121" $\begin{array}{rcl}& & \frac{\mathrm{dw}}{\mathrm{dv}}=2\mathrm{v}\phantom{\rule{0ex}{0ex}}\mathrm{dv}=\frac{1}{2\mathrm{v}}\mathrm{dw}\phantom{\rule{0ex}{0ex}}\end{array}$

Now,

$\begin{array}{rcl}\int \frac{\mathrm{v}}{\sqrt{\mathrm{w}}}\frac{1}{2\mathrm{v}}\mathrm{dw}& =& \frac{1}{2}\int \frac{1}{\sqrt{\mathrm{w}}}\mathrm{dw}\\ & =& \frac{1}{2}\left(2\right)\sqrt{\mathrm{w}}\\ & =& \sqrt{\mathrm{w}}\end{array}$

Substitute $\mathrm{w}=1+{\mathrm{v}}^{2}$.

$\begin{array}{rcl}\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^{2}}}\mathrm{dv}& =& \sqrt{\mathrm{w}}\\ & =& \sqrt{1+{\mathrm{v}}^{2}}\end{array}$

Then,

$\sqrt{1+{\mathrm{v}}^{2}}=\mathrm{ln}\left|\mathrm{t}\right|+\mathrm{C}$

Substitute role="math" localid="1655108419921" $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}$.

$\begin{array}{rcl}\sqrt{1+{\left(\frac{\mathrm{x}}{\mathrm{t}}\right)}^{2}}& =& \mathrm{ln}\left|\mathrm{t}\right|+\mathrm{C}\\ \sqrt{1+\frac{{\mathrm{x}}^{2}}{{\mathrm{t}}^{2}}}& =& \mathrm{ln}\left|\mathrm{t}\right|+\mathrm{C}\end{array}$

Therefore, Homogeneous equation for the given equation is $\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{t}}^{\mathbf{2}}}}{\mathbf{=}}{\mathbf{ln}}\mathbf{|}\mathbf{t}\mathbf{|}{\mathbf{+}}{\mathbf{C}}$.