haganmc

solve diff equation
dy/dx =(sec^2(y))/(1+x^2)

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lgbasallote

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\[\frac{dy}{dx} = \frac{\sec^2 y}{1 + x^2}\]
divide both sides by sec^2 y
\[\implies \frac{dy}{\sec^2 y dx} = \frac{1}{1+x^2}\]
now multiply both sides by dx
\[\implies \frac{dy}{\sec^2 y} = \frac{dx}{1+x^2}\]
now you can integrate both sides
\[\implies \int \frac{dy}{\sec^2 y} = \int \frac{dx}{1+x^2}\]
does that help?

haganmc

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yes. the Integration is the problem i'm having trouble with

lgbasallote

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okay first turn dy/ sec^2 y into cos^2 y dy
you agree those are the same right?

lgbasallote

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\[\implies \int \cos^2 y dy = \int \frac{dx}{1+x^2}\]

haganmc

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yes now i see.

lgbasallote

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wonderful

haganmc

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thanks

lgbasallote

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welcome

haganmc

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what about 1/(1+x^2)

lgbasallote

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trigonometric substitution

haganmc

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is it tan?

lgbasallote

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arctan*

haganmc

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right

lgbasallote

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yes

haganmc

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that's what I thought. Thanks

lgbasallote

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welcome