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Q 2.6-41E

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: Use the substitution ${\mathbf{v}}{\mathbf{=}}{\mathbf{x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{+}}{\mathbf{2}}$ to solve equation (8).$\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}{\mathbf{y}}{\mathbf{-}}{\mathbf{x}}{\mathbf{-}}{\mathbf{1}}{\mathbf{+}}{\left(x-y+2\right)}^{-1}$

${\left(x-y+2\right)}^{2}\mathbf{=}C{\mathbf{e}}^{2x}\mathbf{+}\mathbf{1}$

See the step by step solution

Step 1: Use the given substitution v=x-y+2  to solve the given equation,

The given substitution,

$\mathbf{v}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{+}\mathbf{2}$

Simplify the above equation,

$\begin{array}{l}\mathbf{1}\mathbf{-}\mathbf{v}\mathbf{=}\mathbf{1}\mathbf{-}\left(x-y+2\right)\\ 1-v=y-x-1\end{array}$

And

role="math" localid="1663933729097" $\begin{array}{c}1-v=y-x-1\\ y=1-v+x+1\\ y=x-v+2\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathrm{dv}}{\mathrm{dx}}\end{array}$

Given equation,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{+}{\left(x-y+2\right)}^{-1}$

Substitute $\mathbf{v}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{+}\mathbf{2},\text{\hspace{0.17em}}\mathbf{1}\mathbf{-}\mathbf{v}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{1}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathrm{dv}}{\mathrm{dx}}$ in the above equation,

$\mathbf{1}\mathbf{-}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{v}\mathbf{+}{\left(v\right)}^{-1}$

Simplify the above equation,

$\begin{array}{l}\mathbf{-}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\mathbf{v}\mathbf{+}{\left(v\right)}^{-1}\\ \frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\mathbf{v}\mathbf{-}\frac{1}{v}\\ \frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{v}}^{2}\mathbf{-}\mathbf{1}}{v}\end{array}$

Cross multiplication on both sides in the above equation,

$\begin{array}{l}\frac{v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{1}}\mathbf{dv}\mathbf{=}\mathbf{dx}\\ \frac{2v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{1}}\mathbf{dv}\mathbf{=}\mathbf{2}\text{ }\mathbf{dx}\end{array}$

Step 2: Integrating

Taking integration both sides in the above equation

$\int \frac{2v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{1}}\mathbf{dv}\mathbf{=}\mathbf{2}\int \mathbf{dx}$

Solve the above equation,

$\begin{array}{c}\int \frac{2v}{{\mathbf{v}}^{2}\mathbf{-}\mathbf{1}}\mathbf{dv}\mathbf{=}\mathbf{2}\int \mathbf{dx}\\ \mathbf{log}\left({\mathbf{v}}^{2}\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{log}\text{ }C\\ \frac{{\mathbf{v}}^{2}\mathbf{-}\mathbf{1}}{C}\mathbf{=}{\mathbf{e}}^{2x}\\ {\mathbf{v}}^{2}\mathbf{-}\mathbf{1}\mathbf{=}C\text{ }{\mathbf{e}}^{2x}\\ {\mathbf{v}}^{2}\mathbf{=}C\text{ }{\mathbf{e}}^{2x}\mathbf{+}\mathbf{1}\end{array}$

Substitute the value of $\mathbf{v}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{+}\mathbf{2}$ in the above equation,

${\left(x-y+2\right)}^{2}\mathbf{=}C{\mathbf{e}}^{2x}\mathbf{+}\mathbf{1}$

Hence the solution is ${\left(x-y+2\right)}^{2}{\mathbf{=}}{C}{{\mathbf{e}}}^{2x}{\mathbf{+}}{\mathbf{1}}$