Question

help pls. topic linear differential equation

Answer 1

problem is \[\sin y dy - cosy(1-xcosy) dx = 0\]

Answer 2

@johnweldon1993

Answer 3

and the final answer should be \[secy = x+1 +ce^x\]

Answer 4

no i mean divide by sinydx

Answer 5

so it gives \[\frac{ dy }{ dx } - \frac{ \cos }{ \sin }(\frac{ 1 }{ siny dx }-\frac{ xcos }{ sindx}) = 0\]

Answer 6

hey @baru

Answer 7

this is supposed to be first order linear right,

i'm not able to put it into standard form...

Answer 8

yes it is

Answer 9

can you just double check and see if you have typed the question correctly?
i just cant seem to get it into standard form

Answer 10

yes i typed the question correctly

Answer 11

i think this is non linear,
maybe we can solve by separating variables

Answer 12

oh. lets try by separable

Answer 13

\[\sin y dy - cosy(1-xcosy) dx = 0\\sin(y)dy=\cos(y)(1-xcos(y))dx\\ \frac{\sin(y)}{\cos(y)}dy=(1-xcos(y))dx\]

Answer 14

come to think of it.. we cant separate variables either...

Answer 15

@Kainui

Answer 16

sin/cos is tan

Answer 17

its a difficult problem tho

Answer 18

yep, looks right

Answer 19

what section is the first problem in?

Answer 20

in your book

Answer 21

linear equation in diff equation

Answer 22

but it is not linear as @baru pointed out

Answer 23

can you tell me what methods you have discussed ?

Answer 24

this is the only formula give by our from, i guess its a 1st degree linear diff

Answer 25

so you guys haven't chatted about bernoulli equations ?

Answer 26

@myininaya can u check my work in this problem
problemis \[(x-2) \frac{ dy }{ dx } = y+ 2(x-2)^3\]
divide both side by x-2
\[\frac{ dy }{ dx } = \frac{ y }{ x-2 } +2(x-2)^2\]
\[\frac{ dy }{ dx } - \frac{ y }{ x-2 } = 2(x-2)^2\]
p(x) = -1/x-2 and q(x) = 2(x-2)^2
I.F = \[e ^{\int\limits \frac{ -1 }{ x-2 }} dx\]

Answer 27

that looks right so far

Answer 28

im having trouble in the I.F

Answer 29

\[\int\limits \frac{1}{u} du=\ln|u|+C\]

Answer 30

so just =ln/x-2/

Answer 31

well you have have a negative 1 as constant multiple

\[\int\limits \frac{-1}{x-2} dx=- \int\limits \frac{1}{x-2} dx \\ \text{ \let } u=x-2 \\ \text{ then } du=dx \\ \int\limits \frac{-1}{x-2} dx=\int\limits \frac{-du}{u}=-\ln|u|+C=-\ln|x-2|+C \\ =\ln|(x-2)^{-1}|+C\]

Answer 32

we won't not need the constant of integration though

Answer 33

ah okay, so the solution is
\[\ln(x-2)^{-1} y = 2(x-2)^2 (\ln(x-2)^{-1} dx + c\]

Answer 34

did you put your whole integrating factor in?
because it looks like you only consider the exponent part of the integrating factor

Answer 35

yes, the formula is.
\[(I.F)y = \int\limits Q(x) (I.F) dx + c\]

Answer 36

you did not put in the integrating factor though
you only put in the exponent part of the integrating factor

Answer 37

also the integrating factor can be simplified more

Answer 38

what u mean

Answer 39

you do remember you called I.F. e^(int p dx) ?

Answer 40

yes

Answer 41

so what happen to the e^ part of the I.F. ?

Answer 42

the e part become(x-2)^-1

Answer 43

so you mean e^(ln(x-2)^(-1)) became (x-2)^(-1) or just 1/(x-2)

Answer 44

just 1/ x-2

Answer 45

i was asking this because you put in ln((x-2)^(-1)) as I.F. instead

Answer 46

like either (x-2)^(-1) or 1/(x-2) would have worked as I.F.

Answer 47

ah, sorry i forgot the e

Answer 48

sorry if i made it confusing

Answer 49

\[\frac{ 1 }{ x-2 } y = 2(x-2)^2 (\frac{ 1 }{ x-2 }) dx + c\]

Answer 50

i'm going to go ahead and put in that integral sign
\[\frac{1}{x-2}y= \int\limits 2 (x-2)^2 \frac{1}{x-2} dx +C \\ \frac{1}{x-2} y=\int\limits 2 (x-2) dx+C\]
perform the last calculus step which is the integration part

Answer 51

yes then integrate the \[\int\limits 2(x-2) dx + c\] is
\[2(\frac{ (x-2)^2 }{ 2 }) + c \] to
\[(x-2)^2\]

Answer 52

yah for that one part yes

Answer 53

\[\frac{ 1 }{ x-2 } y = (x-2)^2 + c\]

Answer 54

divide both side by x-2

Answer 55

\[y = (x-2) + c(x-2)\]

Answer 56

but the book answer is \[y = (x-2)^3 + c(x-2)\]

Answer 57

you didn't divide both sides by (x-2) correctly :p

also you should have multiplied both sides by (x-2)

Answer 58

wait im confused

Answer 59

example
if you had
y/5=6
you would solve this by multiplying 5 on both sides
not by dividing both sides by 5

Answer 60

y/5=6
5(y/5)=5(6)
y=30


we did this because we knew 5/5 is 1
and 1*y is y

Answer 61

so same thing here we had y/(x-2) on the left hand side
we knew to multiply both sides by (x-2) because this would get that y by itself

Answer 62

yes i should multiply instead of multiplying

Answer 63

since (x-2)/(x-2)=1

Answer 64

assuming of course x is never 2

Answer 65

same answer with the book

Answer 66

thanks for the help @myininaya

Answer 67

np