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Expert-verified Found in: Page 77 ### 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: Coupled Equations. In analyzing coupled equations of the form$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dt}}\mathbf{=}\mathbf{ax}\mathbf{+}\mathbf{by}\\ \frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{\alpha x}\mathbf{+}\mathbf{\beta y}\end{array}$where a, b, ${\mathbf{\alpha }}{\text{\hspace{0.17em}}}{\mathbf{and}}{\text{\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}$ are constants, we may wish to determine the relationship between x and y rather than the individual solutions x(t), y(t). For this purpose, divide the first equation by the second to obtain $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}\frac{\mathrm{ax}+\mathrm{by}}{\mathrm{\alpha x}+\mathrm{\beta y}}$This new equation is homogeneous, so we can solve it via the substitution ${\mathbf{v}}{\mathbf{=}}\frac{x}{y}$. We refer to the solutions of (17) as integral curves. Determine the integral curves for the system$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dt}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}\\ \frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\end{array}$

${\left(y+x\right)}^{3}{\left(y-4x\right)}^{2}\mathbf{=}C$

See the step by step solution

## Step 1: The given equation is a homogeneous differential equation

The given equations are,

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dt}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}\\ \frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\end{array}$

Divide the first equation by the second equation,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{-4x-y}{2x-y}$

Substitute the $\mathbf{y}\mathbf{=}\mathbf{vx}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}$ in the above equation,

$\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{-4x-\mathrm{vx}}{2x-\mathrm{vx}}$

Simplify the above equation,

$\begin{array}{l}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{-4-v}{2-v}\mathbf{-}\mathbf{v}\\ \mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{-}\mathbf{v}\mathbf{-}\mathbf{2}\mathbf{v}\mathbf{+}{\mathbf{v}}^{2}}{2-v}\\ \mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{v}}^{2}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{4}}{2-v}\end{array}$

Cross multiplication on both sides in the above equation,

$\frac{2-v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{4}}\mathbf{dv}\mathbf{=}\frac{\mathrm{dx}}{x}$

## Step 2: Integrating

Taking integration both sides in the above equation

$\int \frac{2-v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{4}}\mathbf{dv}\mathbf{=}\int \frac{\mathrm{dx}}{x}$

Solve the above equation,

$\begin{array}{l}\int \frac{2-v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{4}}\mathbf{dv}\mathbf{=}\int \frac{\mathrm{dx}}{x}\\ \int \frac{2}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{4}}\mathbf{dv}\mathbf{-}\int \frac{v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{4}}\mathbf{dv}\mathbf{=}\int \frac{\mathrm{dx}}{x}\\ \mathbf{2}\left(\mathbf{-}\frac{1}{5}\mathbf{ln}\left|v+1\right|\mathbf{+}\frac{1}{5}\mathbf{ln}\left|v-4\right|\right)\mathbf{-}\left(\frac{1}{5}\mathbf{ln}\left|v+1\right|\mathbf{+}\frac{4}{5}\mathbf{ln}\left|v-4\right|\right)\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{ln}\left|c\right|\\ \mathbf{-}\frac{3}{5}\mathbf{ln}\left|v+1\right|\mathbf{-}\frac{2}{5}\mathbf{ln}\left|v-4\right|\mathbf{-}\mathbf{ln}\left|x\right|\mathbf{=}\mathbf{ln}\left|c\right|\end{array}$

## Step 3: Finding the solution

Substitute the $\mathbf{v}\mathbf{=}\frac{y}{x}$ in the above equation,

$\begin{array}{l}\mathbf{-}\frac{3}{5}\mathbf{ln}\left|\frac{y}{x}\mathbf{+}\mathbf{1}\right|\mathbf{-}\frac{2}{5}\mathbf{ln}\left|\frac{y}{x}\mathbf{-}\mathbf{4}\right|\mathbf{-}\mathbf{ln}\left|x\right|\mathbf{=}\mathbf{ln}\left|c\right|\\ \mathbf{-}\mathbf{3}\mathbf{ln}\left|\frac{y}{x}\mathbf{+}\mathbf{1}\right|\mathbf{-}\mathbf{2}\mathbf{ln}\left|\frac{y}{x}\mathbf{-}\mathbf{4}\right|\mathbf{-}\mathbf{5}\mathbf{ln}\left|x\right|\mathbf{=}\mathbf{5}\mathbf{ln}\left|c\right|\\ \mathbf{3}\mathbf{ln}\left|\frac{y}{x}\mathbf{+}\mathbf{1}\right|\mathbf{+}\mathbf{2}\mathbf{ln}\left|\frac{y}{x}\mathbf{-}\mathbf{4}\right|\mathbf{+}\mathbf{5}\mathbf{ln}\left|x\right|\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{ln}\left|c\right|\\ {\left(\frac{y}{x}\mathbf{+}\mathbf{1}\right)}^{3}{\left(\frac{y}{x}\mathbf{-}\mathbf{4}\right)}^{2}{\mathbf{x}}^{5}\mathbf{=}{c}^{-5}\end{array}$

Simplify the above equation,

$\begin{array}{c}\frac{{\left(y+x\right)}^{3}{\left(y-4x\right)}^{2}{\mathbf{x}}^{5}}{{\mathbf{x}}^{3}\cdot {\mathbf{x}}^{2}}\mathbf{=}{\mathbf{c}}^{-5}\\ {\left(y+x\right)}^{3}{\left(y-4x\right)}^{2}\mathbf{=}{\mathbf{c}}^{-5}\end{array}$

Substitute ${\mathbf{c}}^{-5}\mathbf{=}C$ in above equation,

${\left(y+x\right)}^{3}{\left(y-4x\right)}^{2}\mathbf{=}C$

Hence the solution is ${\left(y+x\right)}^{3}{\left(y-4x\right)}^{2}{\mathbf{=}}{C}$ ### Want to see more solutions like these? 