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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

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# Question: In Problems 1 - 30, solve the equation.${\text{\hspace{0.17em}}}\left[\mathbf{1}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^{2}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^{2}\right)}^{-1}\right]{\mathbf{dx}}{\mathbf{+}}\left[{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^{2}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^{2}\right)}^{-1}\right]{\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

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

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 x0 at 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:

• Integration by parts:. $\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$
• $\int {\mathbf{x}}^{a}\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$.

## Step 2: Given information and simplification

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

Evaluate it.

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

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

Then, $\mathbf{M}\mathbf{=}\frac{\mathbf{2}\mathbf{+}{\left(x+y\right)}^{2}}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\mathbf{,}\mathbf{N}\mathbf{=}\frac{{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}1\mathbf{+}{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}{\left(x+y\right)}^{2}}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}$

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\mathbf{2}\left(x+y\right)}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\frac{\mathbf{2}\left(x+y\right)}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\end{array}$

So, the given equation is exact.

## Step 3: Evaluation method

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{1}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\right)\mathbf{dx}\\ \mathbf{=}{\mathbf{tan}}^{-1}\left(x+y\right)\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Differentiate the F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalise the values of N.

$\begin{array}{c}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^{2}}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\end{array}$

Integrate on both sides.

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

Substitute in equation of F.

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

Hence, the solution of the given initial value problem is ${{\mathbf{tan}}}^{-1}\left(x+y\right){\mathbf{+}}{\mathbf{x}}{\mathbf{+}}{\mathbf{2}}\sqrt{y}{\mathbf{=}}{\mathbf{C}}$.

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