anonymous
  • anonymous
solve diff equation dy/dx =(sec^2(y))/(1+x^2)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lgbasallote
  • lgbasallote
\[\frac{dy}{dx} = \frac{\sec^2 y}{1 + x^2}\] 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?
anonymous
  • anonymous
yes. the Integration is the problem i'm having trouble with
lgbasallote
  • lgbasallote
okay first turn dy/ sec^2 y into cos^2 y dy you agree those are the same right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

lgbasallote
  • lgbasallote
\[\implies \int \cos^2 y dy = \int \frac{dx}{1+x^2}\]
anonymous
  • anonymous
yes now i see.
lgbasallote
  • lgbasallote
wonderful
anonymous
  • anonymous
thanks
lgbasallote
  • lgbasallote
welcome
anonymous
  • anonymous
what about 1/(1+x^2)
lgbasallote
  • lgbasallote
trigonometric substitution
anonymous
  • anonymous
is it tan?
lgbasallote
  • lgbasallote
arctan*
anonymous
  • anonymous
right
lgbasallote
  • lgbasallote
yes
anonymous
  • anonymous
that's what I thought. Thanks
lgbasallote
  • lgbasallote
welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.