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Expert-verified Found in: Page 79 ### 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: In Problems 1-30, solve the equation.${\text{\hspace{0.17em}}}{\mathbf{y}}\left(x-y-2\right){\mathbf{dx}}{\mathbf{+}}{\mathbf{x}}\left(y-x+4\right){\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

${\mathbf{x}}^{2}{\mathbf{y}}^{-2}\mathbf{-}\mathbf{2}{\mathbf{xy}}^{-1}\mathbf{-}\mathbf{4}{\mathbf{xy}}^{-2}\mathbf{=}\mathbf{C}\text{ and }\mathbf{y}\equiv \mathbf{0}$

See the step by step solution

## Step 1: Definition and concepts to be used

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}...\mathbf{,}\frac{{\mathbf{d}}^{n}\mathbf{y}}{{\mathbf{dx}}^{n}}\right)\mathbf{=}\mathbf{0}$ we mean: Find a solution to the differential equation on an interval I that satisfies at x0 the n initial conditions

$\begin{array}{c}\mathbf{y}\left({\mathbf{x}}_{0}\right)\mathbf{=}{\mathbf{y}}_{0}\mathbf{,}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathbf{x}}_{0}\right)\mathbf{=}{\mathbf{y}}_{1}\mathbf{,}\\ .\\ .\\ .\\ \frac{{\mathbf{d}}^{n-1}\mathbf{y}}{{\mathbf{dx}}^{n-1}}\left({\mathbf{x}}_{0}\right)\mathbf{=}{\mathbf{y}}_{n-1}\mathbf{,}\mathbf{,}\end{array}$

Where ${\mathbf{x}}_{0}\in \mathbf{I}$ and ${\mathbf{y}}_{0}\mathbf{,}{\mathbf{y}}_{1}\mathbf{,}...\mathbf{,}{\mathbf{y}}_{n-1}$ are given constants.

Exactness: If $\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$. Otherwise, the equation is not exact.

Special integrating factor: $\frac{\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}}{N},\text{ }\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}$.

Formulae to be used:

• $\int {\mathbf{x}}^{a}\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int {\mathbf{e}}^{\mathrm{ax}}\mathbf{dx}\mathbf{=}\frac{{\mathbf{e}}^{\mathrm{ax}}}{a}\mathbf{+}\mathbf{C}$.

## Step 2: Given information and simplification

Given that, $\mathbf{y}\left(x-y-2\right)\mathbf{dx}\mathbf{+}\mathbf{x}\left(y-x+4\right)\mathbf{dy}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}}......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Let us check whether the given equation is exact or not.

Then, $\mathbf{M}\mathbf{=}\mathbf{y}\left(x-y-2\right)\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{x}\left(y-x+4\right)$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{2}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{4}\end{array}$

So, the given equation is not exact. Then, find the special integrating factor,

$\begin{array}{c}\frac{\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}}{M}\mathbf{=}\frac{y-2x+4-x+2y+2}{\mathbf{y}\left(x-y-2\right)}\\ \mathbf{=}\frac{3y-3x+6}{\mathbf{y}\left(x-y-2\right)}\\ \mathbf{=}\frac{\mathbf{-}\mathbf{3}\left(x-y-2\right)}{\mathbf{y}\left(x-y-2\right)}\\ \mathbf{=}\mathbf{-}\frac{3}{y}\end{array}$

## Step 3: Find the exactness

Find the value of $\mu \left(y\right)$.

$\begin{array}{c}\mu \left(y\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{-}\frac{3}{y}\mathbf{dy}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{ln}\left|y\right|}\\ \mathbf{=}{\mathbf{y}}^{-3}\end{array}$

Multiply y--3 in equation (1) on both sides.

${\mathbf{y}}^{-2}\left(x-y-2\right)\mathbf{dx}\mathbf{+}{\mathbf{xy}}^{-3}\left(y-x+4\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Now again check whether the founded equation is exact or not.

$\mathbf{M}\mathbf{=}{\mathbf{y}}^{-2}\left(x-y-2\right)\mathbf{,}\mathbf{N}\mathbf{=}{\mathbf{xy}}^{-3}\left(y-x+4\right)$

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{4}{{\mathbf{y}}^{3}}\mathbf{+}\frac{1}{{\mathbf{y}}^{2}}\mathbf{-}\frac{2x}{{\mathbf{y}}^{3}}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\frac{4}{{\mathbf{y}}^{3}}\mathbf{+}\frac{1}{{\mathbf{y}}^{2}}\mathbf{-}\frac{2x}{{\mathbf{y}}^{3}}\end{array}$

Therefore, the founded equation is exact.

## Step 4: Evaluation method

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}{\mathbf{y}}^{-2}\left(x-y-2\right)$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left({\mathbf{y}}^{-2}\left(x-y-2\right)\right)\mathbf{dx}\\ \mathbf{=}\frac{{\mathbf{x}}^{2}}{\mathbf{2}{\mathbf{y}}^{2}}\mathbf{-}\frac{x}{y}\mathbf{-}\frac{2x}{{\mathbf{y}}^{2}}\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Differentiate the F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\frac{4x}{{\mathbf{y}}^{3}}\mathbf{-}\frac{{\mathbf{x}}^{2}}{{\mathbf{y}}^{3}}\mathbf{+}\frac{x}{{\mathbf{y}}^{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalize the values of N.

$\begin{array}{c}\frac{4x}{{\mathbf{y}}^{3}}\mathbf{-}\frac{{\mathbf{x}}^{2}}{{\mathbf{y}}^{3}}\mathbf{+}\frac{x}{{\mathbf{y}}^{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\frac{4x}{{\mathbf{y}}^{3}}\mathbf{-}\frac{{\mathbf{x}}^{2}}{{\mathbf{y}}^{3}}\mathbf{+}\frac{x}{{\mathbf{y}}^{2}}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}0\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\int 0\text{ }\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}{\mathbf{C}}_{1}\end{array}$

Substitute in the equation of F.

$\frac{{\mathbf{x}}^{2}}{\mathbf{2}{\mathbf{y}}^{2}}\mathbf{-}\frac{x}{y}\mathbf{-}\frac{2x}{{\mathbf{y}}^{2}}\mathbf{+}{\mathbf{C}}_{1}\mathbf{=}\mathbf{0}$

Multiply 2 on both sides.

$\begin{array}{c}{\mathbf{x}}^{2}{\mathbf{y}}^{-2}\mathbf{-}\mathbf{2}{\mathbf{xy}}^{-1}\mathbf{-}\mathbf{4}{\mathbf{xy}}^{2}\mathbf{+}{\mathbf{C}}_{2}\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^{2}{\mathbf{y}}^{-2}\mathbf{-}\mathbf{2}{\mathbf{xy}}^{-1}\mathbf{-}\mathbf{4}{\mathbf{xy}}^{2}\mathbf{=}\mathbf{C}\end{array}$

Hence, the solution of the given initial value problem is ${{\mathbf{x}}}^{2}{{\mathbf{y}}}^{-2}{\mathbf{-}}{\mathbf{2}}{{\mathbf{xy}}}^{-1}{\mathbf{-}}{\mathbf{4}}{{\mathbf{xy}}}^{2}{\mathbf{=}}{\mathbf{C}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{\mathbf{y}}{\mathbf{\equiv }}{\mathbf{0}}$ ### Want to see more solutions like these? 