Question

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

Short answer

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

Step-by-step solution

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.

Solution of the Equation

The given equation is

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

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

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

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

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

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

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