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Q2.2-12E

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Found in: Page 46

### 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

# In problem 7-16, solve the equation. $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}\frac{{\mathbf{sec}}^{2}\mathbf{y}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{2}}$

The solution of the differential equation is ${\mathbf{2}}{\mathbf{y}}{\mathbf{+}}{\mathbf{sin}}{\text{ }}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{\mathbf{4}}{\text{ }}{{\mathbf{tan}}}^{-1}{\mathbf{x}}{\mathbf{+}}{\mathbf{C}}$

See the step by step solution

## Step 1: Concept of Separable Differential Equation

A first-order ordinary differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}{\mathbf{f}}\left(x,y\right)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions ${\mathbf{g}}\left(x\right)$ that is a function of x alone and ${\mathbf{h}}\left(y\right)$that is a function of y alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathbf{h}\left(y\right)}{\mathbf{=}}{\mathbf{g}}\left(x\right){\mathbf{dx}}$. By separating the variables, the equation follows. Then, on direct integration of both sides, the solution of the differential equation is determined.

## Step 2: Solution of the Equation

The given equation is

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

After separating the variables, equation (1) can be written as

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

Integrate both sides of equation (2). It results,

$\begin{array}{c}\int \frac{\mathrm{dy}}{{\mathrm{sec}}^{2}\mathrm{y}}=\int \frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}\\ \int {\mathrm{cos}}^{2}\mathrm{y}\text{ }\mathrm{dy}=\int \frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}\\ \frac{1}{2}\int 2{\mathrm{cos}}^{2}\mathrm{y}\text{ }\mathrm{dy}=\int \frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}\\ \frac{1}{2}\int \left(1+\mathrm{cos}2\mathrm{y}\right)\mathrm{dy}=\int \frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}\left[\mathrm{cos}2\mathrm{y}=2{\mathrm{cos}}^{2}\mathrm{y}-1\right]\end{array}$

$\begin{array}{c}\frac{1}{2}\left[\int \mathrm{dy}+\int \mathrm{cos}2\mathrm{y}\text{ }\mathrm{dy}\right]=\int \frac{dx}{1+{\mathrm{x}}^{2}}\\ \frac{1}{2}\left[\mathrm{y}+\frac{1}{2}\mathrm{sin}2\mathrm{y}\right]={\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{k}\left[\mathrm{k}=\mathrm{Integrating}\mathrm{Constant}\right]\\ 2\mathrm{y}+\mathrm{sin}2\mathrm{y}=4{\mathrm{tan}}^{-1}\mathrm{x}+4\mathrm{k}\\ 2\mathrm{y}+\mathrm{sin}2\mathrm{y}=4{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\left[\mathrm{C}=4\mathrm{k}=\mathrm{Constant}\right]\end{array}$

Therefore, the solution of the given equation is ${\mathbf{2}}{\mathbf{y}}{\mathbf{+}}{\mathbf{sin}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{\mathbf{4}}{{\mathbf{tan}}}^{-1}{\mathbf{x}}{\mathbf{+}}{\mathbf{C}}$.

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