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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.$\left(y-x\right){\mathbf{dx}}{\mathbf{+}}\left(x+y\right){\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

${\mathbf{y}}^{2}\mathbf{+}2\mathbf{xy}\mathbf{-}{\mathbf{x}}^{2}\mathbf{=}\mathbf{C}$

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.

Formulae to be used:

• Exactness: If $\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$. Otherwise, the equation is not exact.
• $\int {\mathbf{x}}^{a}\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$

## Step 2: Given information and simplification

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

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

Then, $\mathbf{M}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}1\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}1\end{array}$

So, the given equation is exact.

## Step 3: Evaluation method

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{x}$.

Integrate on both sides.

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

Differentiate the F with respect to y.

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

Equalise the values of N.

$\begin{array}{c}\mathbf{x}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{y}\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\int \mathbf{ydy}\\ \mathbf{g}\left(y\right)\mathbf{=}\frac{{\mathbf{y}}^{2}}{2}\mathbf{+}{\mathbf{C}}_{1}\end{array}$

Substitute in equation of F.

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

Hence, the solution of the given initial value problem is ${{\mathbf{y}}}^{2}{\mathbf{+}}{2}{\mathbf{xy}}{\mathbf{-}}{{\mathbf{x}}}^{{2}}{\mathbf{=}}{\mathbf{C}}$.