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Q2.6 - 13E

Expert-verified
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 $\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{t}}^{\mathbf{2}}}}{\mathbf{=}}{\mathbf{ln}}\left|\mathbf{t}\right|{\mathbf{+}}{\mathbf{C}}$

See the step by step solution

## General form of Homogeneous equation

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

## Evaluate the given equation

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

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

## Substitution method

Let us take $v=\frac{y}{x}$

Then $y=vx$

By Differentiating,

$\begin{array}{rcl}\frac{\mathbf{dx}}{\mathbf{dt}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ \left(\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{+}\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ & & \end{array}$

$\begin{array}{rcl}\left(\mathbf{v}\mathbf{+}\frac{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ \frac{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}& \mathbf{=}& \mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ \frac{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}\frac{\mathbf{1}}{\mathbf{dv}}& \mathbf{=}& \mathbf{t}\frac{\mathbf{1}}{\mathbf{dt}}\\ \frac{\mathbf{v}}{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}\mathbf{dv}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{t}}\mathbf{dt}\\ & & \end{array}$

## Integrate the equation

Now, integrate on both sides.

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

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

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

Then, $\frac{\mathbf{dw}}{\mathbf{dv}}\mathbf{=}\mathbf{2}\mathbf{v}\phantom{\rule{0ex}{0ex}}\mathbf{dv}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{v}}\mathbf{dw}\phantom{\rule{0ex}{0ex}}$

Now,

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

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

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

Then,

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

Substitute $\mathbf{v}\mathbf{=}\frac{\mathbf{y}}{\mathbf{x}}$

$\sqrt{\mathbf{1}\mathbf{+}{\left(\frac{\mathbf{x}}{\mathbf{t}}\right)}^{\mathbf{2}}}\mathbf{=}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}\mathbf{C}\phantom{\rule{0ex}{0ex}}\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{t}}^{\mathbf{2}}}}\mathbf{=}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}\mathbf{C}$

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