Question

help with diff eqns

Answer 1

\[2\sqrt{x}\frac{ dy }{ dx }=\cos^2(y), y(4) = \pi/4\]

Answer 2

\[2 \sec^2(y) dy =x^\frac{-1}{2} dx\]
integrate both sides

Answer 3

\[2\tan(y)= 2\sqrt{x}+C\] ?

Answer 4

I think that looks good

Answer 5

now you are given x=4 and y=pi/4 find C

Answer 6

C=-1?

Answer 7

\[2(1)=2(2)+C\\ 2=4+C\]

Answer 8

I got that originally

Answer 9

but I went back and simplified the general solution

Answer 10

that gives \[y = \tan^{-1} (\sqrt{x}+C)\]

Answer 11

so \[\frac{ \pi }{ 4 }= \tan^{-1} (2+C)\]

Answer 12

\[\tan (\frac{ \pi }{ 4 })= 2+C\]

Answer 13

\[1=2+C\]
\[C=-1\]

Answer 14

\[\tan(y)=\sqrt{x}+k \\ y=\arctan(\sqrt{x}+k) \\ \frac{\pi}{4}=\arctan(\sqrt{4}+k) \\ \frac{\pi}{4}=\arctan(2+k) \\ 1=2+k \\k=-1 \\ \tan(y)=\sqrt{x}-1 \]
but this will give you the same thing if you used the other C from before:
\[2 \tan(x)=2 \sqrt{x}+C \\ 2 \tan(x)=2 \sqrt{x}-2 \\ \text{ divide both sides by 2} \\ \tan(y)=\sqrt{x}-1 \]

Answer 15

I didn't know you called C something different

Answer 16

something different? C is just a constant

Answer 17

are we saying the same thing?

Answer 18

I'm saying your constant is correct
I just didn't know you called it something different than what we had before

Answer 19

ohhh!

Answer 20

so particular solution is \[y=\tan^{-1} (\sqrt{x}-1)\]

Answer 21

?

Answer 22

yeah

Answer 23

ok, id like to try another, can you correct as I go?

Answer 24

yep

Answer 25

\[y'=(1-y)cosx, y(\pi)=2\]

Answer 26

and you said you are going to show work
and let me check it?

Answer 27

yes

Answer 28

ok

Answer 29

\[\frac{ y' }{ 1-y }=cosx\]
\[\int\limits_{}^{}\frac{ 1 }{ 1-y }dy=\int\limits_{}^{}cosxdx\]
\[-\ln(1-y)=sinx+C\]

Answer 30

awesome so far

Answer 31

thought

Answer 32

one correction

Answer 33

-ln|1-y|=sin(x)+C

Answer 34

if you apply the condition as is you will run into some domain problems with the other equation

Answer 35

solve explicitly for y?

Answer 36

it seems to be doable so yep

Answer 37

I normally apply the condition before solving for y

Answer 38

but either way should work

Answer 39

ok, general solution is \[y=e ^{-(sinx+C)}+1\]

Answer 40

a lot of people like to write e^C as just K
\[y=e^{-\sin(x)-C}+1 \\ y=e^{-\sin(x)}e^{-C}+1 \\ y=K e^{-\sin(x)}+1\]

Answer 41

e^c or e^(-c)
whatever it is
it is still just a constant

Answer 42

ohh o, so C=1?

Answer 43

as in you are calling my K ,1 right?

Answer 44

yes

Answer 45

because sometimes like before depending on what C is called you get a different answer
like if you left C in the exponent area it would have been 0

Answer 46

and of course e^0 is 1

Answer 47

\[y=Ke^{-\sin(x)}+1 \\ y=1e^{-\sin(x)}+1 \\y=e^{-\sin(x)}+1\]
sounds perfect ! :)

Answer 48

phewww

Answer 49

lol you did it!

Answer 50

I still think the harder one was the first one

Answer 51

lol, am I annoying you yet?

Answer 52

I didn't see the grouping thing at first

Answer 53

no

Answer 54

but I do have to leave you
and it was totally fun (not annoying)

Answer 55

lol, thank you so much!