What is a solution to the differential equation $\frac{d y}{d x} = \frac{1}{{sec}^{2} y}$ $dy/dx=1/sec^2y$ ?

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Answer 1 · 2016-10-20T21:56:19

$y = {tan}^{- 1} (x + C)$ $y=tan^(-1)(x+C)$

$\frac{d y}{d x} = \frac{1}{{sec}^{2} y}$ $dy/dx = 1/sec^2y$

This is a first order separable Differential Equation; so we can "separate the variables" to give:

$\int {sec}^{2} y d y = \int 1 d x$ $int sec^2ydy = int1dx$

Integrating gives:

$tan y = x + C$ $tan y = x + C$ , where $C$ $C$ is the constant of integration
Hence, the solution is:

$y = {tan}^{- 1} (x + C)$ $y=tan^(-1)(x+C)$

Answer 2 · 2016-10-20T21:58:03

$y = arctan (x + C)$ $y = arctan(x+C)$

Grouping variables

$\frac{d y}{{cos}^{2} y} = d x$ $dy/cos^2y=dx$ so

$tan y = x + C$ $tan y = x + C$ so

$y = arctan (x + C)$ $y = arctan(x+C)$

What is a solution to the differential equation dy/dx=1/sec^2ydydx=1sec2ydy/dx=1/sec^2y?