Question

Solve (sec^2 y)y'-3tan(y)=-1.

Answer 1

rewrite it as tany first

Answer 2

What do you mean? How?

Answer 3

oh hold on! add 3tany to both sides
and then divide by sec^2y

Answer 4

Substitute \(u=\tan y\), then \(u'=\sec^2y~y'\). The DE then changes to
\[u'-3u=-1\]

Answer 5

it is a lot easier the way siths did it!

Answer 6

Or treat it as a separable equation:
\[\begin{align*}
\sec^2y~y'&=3\tan y-1\\
\frac{\sec^2y}{3\tan y-1}~dy&=dx
\end{align*}\]
Then integrate with a substitution, \(u=3\tan y-1\), so \(\dfrac{1}{3}du=\sec^2y~dy\):
\[\frac{du}{3u}=dx\]

Answer 7

But what's the next step of u'-3u=-1?

Answer 8

that's what i was thinking about siths^_^

Answer 9

have you ever done DE?

Answer 10

Yes.

Answer 11

Well my first suggested method changes the equation to a linear equation, so you can find the integrating factor and so on.

The second method is simpler (I guess you could call it that) because separable equations are generally easy to solve.

Answer 12

So du/3u=dx after integrating, you would get ln abs(3u)=x, right?

Answer 13

+C, yes

Answer 14

Got it! Thanks!

Answer 15

yw