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How do you find the general solution to #dy/dx=1/sec^2y#?

Answer 1

Luke Alvarez

# y = tan^(-1) (x + C)#

we can separate it

#dy/dx=1/sec^2y#

#sec^2y dy/dx=1#

#int \ sec^2y dy/dx \ dx =int \ dx#

#int \d/dx( tan y) \ dx =int \ dx#

# tan y =x + C#

# y = tan^(-1) (x + C)#

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Answer 2

Isaiah Allen

To find the general solution to the differential equation (\frac{dy}{dx} = \frac{1}{\sec^2 y}), you would follow these steps:

Rewrite the equation using trigonometric identities: (\frac{dy}{dx} = \cos^2 y).
Separate variables: (\frac{dy}{\cos^2 y} = dx).
Integrate both sides: (\int \frac{dy}{\cos^2 y} = \int dx).
The integral of (\frac{1}{\cos^2 y}) with respect to (y) is (\tan y): (\tan y = x + C), where (C) is the constant of integration.

So, the general solution to the differential equation is: [ \tan y = x + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.