Question

solve

Answer 1

\[y'+y^2 \tan(t)=0, y(0)=1\]

Answer 2

@AccessDenied

Answer 3

Have you tried anything with this so far, or not sure where to begin?

Answer 4

not sure where to begin?

Answer 5

Ah, okay.
When I first looked at the problem, the y^2 definitely struck me as off because the equation was made nonlinear.
I think the key point here though is that the right hand side is zero, and the only term other than y' is a product of a function of y and function of t. That would be separable!

We could subtract the y^2 tan(t) from both sides, and then separate the variables with y'. Do you know how to do that work?

Answer 6

\[y'=-y^2 \tan(t)\]

Answer 7

Yup, can you see then with \( y' = \dfrac{dy}{dt} \) how that can separate out?

Answer 8

\[\frac{ dy }{dt }=-y^2 \tan(t)\]

Answer 9

\[dy(-y^2)=\tan(t) dt\]

Answer 10

But, I dont think this is right

Answer 11

Almost; by moving the \(y^2\) over, you have to divide it off rather than multiply:
\( \dfrac{dy}{dt} = -y^2 \tan (t) \)
\( \dfrac{1}{-y^2} \ dy = \tan (t) \ dt \)
For integration purposes, I might even write it using a negative exponent as well:
\( -y^{-2} \ dy = \tan (t) \ dt \)

Answer 12

okay, got it. Thanks

Answer 13

Glad to help! :)