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Q2.6 - 12E

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. ${\mathbf{ }}\left({x}^{2}+{y}^{2}\right){\mathbf{ }}{\mathbit{d}}{\mathbit{x}}{\mathbf{+}}{\mathbf{2}}{\mathbit{x}}{\mathbit{y}}{\mathbf{}}{\mathbit{d}}{\mathbit{y}}{\mathbf{=}}{\mathbf{0}}$

Homogeneous equation for the given equation is ${{\mathbf{x}}}^{{\mathbf{3}}}{\mathbf{+}}{\mathbf{3}}{{\mathbf{xy}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{C}}$.

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 $\frac{\mathbf{y}}{\mathbf{x}}$ alone, then we say the equation is homogeneous.

## Evaluate the given equation

Given, $\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\right) \mathbf{dx}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{dy}\mathbf{=}\mathbf{0}$

Evaluate it.

$\begin{array}{rcl}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\right) \mathbf{dx}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{dy}& \mathbf{=}& \mathbf{0}\\ \mathbf{2}\mathbf{xydy}& \mathbf{=}& \mathbf{-}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{dx}\\ \frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{\mathbf{2}\mathbf{xy}}\\ & \mathbf{=}& \mathbf{-}\frac{\mathbf{x}}{\mathbf{2}\mathbf{y}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{2}\mathbf{x}}\\ & \mathbf{=}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(\frac{\mathbf{x}}{\mathbf{y}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\right)\\ & & \end{array}$

## Substitution method

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

Then $\mathbf{y}\mathbf{=}\mathbf{vx}$

By Differentiating,

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(\frac{\mathbf{1}}{\mathbf{v}}\mathbf{+}\mathbf{v}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(\frac{\mathbf{1}}{\mathbf{v}}\mathbf{+}\mathbf{v}\right)\mathbf{-}\mathbf{v}& \mathbf{=}& \mathbf{x}\frac{\mathbf{dv}}{\mathbf{dx}}\\ \mathbf{-}\frac{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}\mathbf{v}}\mathbf{-}\mathbf{vdv}& \mathbf{=}& \mathbf{xdx}\\ & & \end{array}$

## Integrate the equation

Now, integrate on both sides.

$\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\mathbf{-}\int \frac{\mathbf{1}}{\mathbf{x}}\mathbf{dx}\phantom{\rule{0ex}{0ex}}\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}\mathbf{=}\mathbf{-}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}\mathbf{C}$

Integrate$\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}$ separately.

Let us take $\mathbf{w}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}$

Then,

$\frac{\mathbf{dw}}{\mathbf{dv}}\mathbf{=}\mathbf{6}\mathbf{v}\phantom{\rule{0ex}{0ex}}\mathbf{dv}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}\mathbf{v}}\mathbf{dw}\phantom{\rule{0ex}{0ex}}$

Now,

$\begin{array}{rcl}\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{w}}\frac{\mathbf{1}}{\mathbf{6}\mathbf{v}}\mathbf{dw}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\int \frac{\mathbf{1}}{\mathbf{w}}\mathbf{dw}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{w}\right|\\ & & \end{array}$

Substitute $\mathbf{w}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}$

$\begin{array}{rcl}\int \frac{\mathbf{2}\mathbf{v}}{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}}\mathbf{dv}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{w}\right|\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right|\\ & & \end{array}$

Then,

$\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right|& \mathbf{=}& \mathbf{-}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \frac{\mathbf{1}}{\mathbf{3}}\mathbf{ln}\left|\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right|\mathbf{+}\mathbf{ln}\left|\mathbf{x}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{1}}\\ \mathbf{ln}\left|\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right|\mathbf{+}\mathbf{3}\mathbf{ln}\left|\mathbf{x}\right|& \mathbf{=}& \mathbf{3}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{ln}\left|\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right|\mathbf{+}\mathbf{ln}\left|{\mathbf{x}}^{\mathbf{3}}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}} \\ & & \end{array}$

$\begin{array}{rcl}\mathbf{ln}\left|\left(\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right)\mathbf{×}{\mathbf{x}}^{\mathbf{3}}\right|& \mathbf{=}& {\mathbf{C}}_{\mathbf{2}}\\ \left(\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right)\mathbf{×}{\mathbf{x}}^{\mathbf{3}}& \mathbf{=}& {\mathbf{e}}^{{\mathbf{C}}_{\mathbf{2}}}\\ \left(\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{v}}^{\mathbf{2}}\right)\mathbf{×}{\mathbf{x}}^{\mathbf{3}}& \mathbf{=}& \mathbf{C}\\ & & \end{array}$

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

$\begin{array}{rcl}{\mathbf{x}}^{\mathbf{3}}\left(\mathbf{1}\mathbf{+}\mathbf{3}{\left(\frac{\mathbf{y}}{\mathbf{x}}\right)}^{\mathbf{2}}\right)& \mathbf{=}& \mathbf{C}\\ {\mathbf{x}}^{\mathbf{3}}\left(\mathbf{1}\mathbf{+}\mathbf{3}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\right)& \mathbf{=}& \mathbf{C}\\ {\mathbf{x}}^{\mathbf{3}}\left(\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\right)& \mathbf{=}& \mathbf{C}\\ \mathbf{x}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\right)& \mathbf{=}& \mathbf{C}\\ {\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{xy}}^{\mathbf{2}}& \mathbf{=}& \mathbf{C}\\ & & \end{array}$

Therefore, Homogeneous equation for the given equation is ${{\mathbf{x}}}^{{\mathbf{3}}}{\mathbf{+}}{\mathbf{3}}{{\mathbf{xy}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{C}}$.