Question

In Problems 45-50 use a technique of integration or a substitution to find an explicit solution of the given differential equation or initial-value problem.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{1}{1+\mathrm{sinx}}$

Short answer

The explicit solution is $y=\tan x-\sec x+c$.

Step-by-step solution

Definition

A first-order differential equation of the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{h}\mathbf{\left(}\mathbf{y}\mathbf{\right)}$is said to be separable or to have separable variables.

Separate variable

Consider the differential equation$\frac{dy}{dx}=\frac{1}{1+\sin x}$

Separate variables by multiplyingto the right hand side.

$\begin{array}{c}\mathrm{dx}\left(\frac{dy}{dx}\right)=\left(\frac{1}{1+\sin x}\right)dx\\ dy=\frac{1}{1+\sin x}dx\end{array}$

Integrating

Now we can integrate the left hand side in terms of $y$and the right hand side in terms of $x$.

$\int dy=\int \frac{1}{1+\sin x}dx$

In order to solve the right hand side integral we can multiply by a convenient form of $1$.

$\begin{array}{c}\frac{1}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{1-{\sin }^{2}x}\\ =\frac{1-\sin x}{{\cos }^{2}x}\\ =\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}\\ ={\sec }^{2}x-\tan x\sec x\end{array}$

The integral equation becomes

$\begin{array}{c}\int dy=\int ({\sec }^{2}x-\tan x\sec x)dx\\ y=\tan x-\sec x+c\end{array}$

Therefore,$y=\tan x-\sec x+c$ .