Question

solve ysec2x y'=-x(1-y^2)

Answer 1

I'm going to assume this is an IVP? Correct?

Answer 2

\[y \sec2x \frac{ dy }{ dx}=-x \sqrt{1-y ^{2})}\]

Answer 3

Separate the x's and y's on each side and take the integral of each

Answer 4

\[\int\limits \frac{y }{ 1+y^2} dy = \int\limits \frac{ x }{ \sec(2x)}dx\]

Answer 5

If you aren't given any initial conditions, then you can leave your answer in the form
f(y) = g(x) + c

Answer 6

why did the 1-y^2 turn to 1+y^2

Answer 7

Typo. My bad homie
It should be: 1-y^2

Answer 8

i am coming up with \[y=\sqrt{1-(\frac{ x }{2}}\sin 2x +1/4\cos2x+c)^{2}\]

Answer 9

would you think that is realistic?

Answer 10

no.

Answer 11

Is this the problem you are solving for:
\[y \sec(2x)\frac{ dy }{ dx } = x (1-y^2)\]
correct?

Answer 12

What you have there is a nonlinear first order differential equation. and since you aren't given any initial conditions, you can leave your answer in implicit form.

Answer 13

i am given \[y(\pi)=1\]

Answer 14

That means that y = 1 and x = π
So just substitute and solve for 'C'
once you have your C, plug it back into your general solution to get your explicit solution.
Did you figure out your general solution?

Answer 15

what i typed ealier is what i am getting for y

Answer 16

i am getting c as -1/4

Answer 17

I don't see how you got that. Your implicit solution should be:
\[\frac{ 1 }{ 2 }\ln(y^2-1) = -[ xsin(x)+\cos(x)+c] \]

Answer 18

when i separate them this is what i got\[\int\limits_{}^{}\frac{ y }{\sqrt{(1-y^2}}dy=\int\limits_{}^{}xcos2xdx\]

Answer 19

I didn't know there was a square root. My answer is wrong then Let me re-work it. Give me a minute.

Answer 20

but that looks correct. the way you separated them!

Answer 21

Solution:
\[-\sqrt{1-y^2}= \frac{ 1 }{ 4 }(-2xsin(2x)-\cos(2x))\]

Answer 22

C = -1

Answer 23

we can multiply by -1 through out? and you forgot the c

Answer 24

Yeah. I 4got to add +C. Good catch!
So your solution should be:
\[-\sqrt{1-y^2} = \frac{ 1 }{ 4 }[-2xsin(2x)-\cos(2x)-1]\]

Answer 25

this is what i got before i further simplified

Answer 26

\[\sqrt{1-y ^{2}}=\frac{ x }{ 2 }\sin2x +\frac{ 1 }{ 4 }\cos2x+c\]

Answer 27

That's correct.

Answer 28

sub x=pi and y=1 RHS=0 sin2x term becomes zero leaving 1/4+c=0

Answer 29

You're correct. I 4got abt the 1/4. My bad. Haha. But you get it better than me, since u caught my mistakes :)

Answer 30

no, it took a while to do it and u was doin it as we were talkin so u r better

Answer 31

and u juss confirmed my answers. thx

Answer 32

no prb. best of luck.