Question

help me in linear differential equation pls

Answer 1

PROBLEM IS \[ydx +(xy+x-3y) dy = 0\]

Answer 2

@ganeshie8

Answer 3

Rewrite as
\[\dfrac{d\color{red}{x}}{dy} + \dfrac{y+1}{y}\color{red}{x} = 3\]

Answer 4

so divide both side by y dy

Answer 5

\[\frac{ dx }{ dy }+ y(xy+x-3y) = 0\]

Answer 6

@ganeshie8

Answer 7

looks you have multiplied y to second term instead of dividing

Answer 8

oops yes \[\frac{ dx }{ dy } +x +\frac{ x }{ y }-3\]

Answer 9

put it in standard form

Answer 10

\[\frac{ dx }{ dy } + (\frac{ xy+x }{ y })\]

Answer 11

= 3

Answer 12

keep going

Answer 13

then the p(y) = (xy+x/y)(x) and Q is 3

Answer 14

No

Answer 15

why

Answer 16

first put the equation in standard form so that you don't make mistakes like above

Answer 17

i already did

Answer 18

You haven't

Answer 19

what is the standard form ?

Answer 20

ax+ by =c

Answer 21

no

Answer 22

why

Answer 23

standard form :
\[\dfrac{dx}{dy} + P(y)*y=Q(y)\]

Answer 24

ah standard form in linear

Answer 25

the standard form for x given by our prof is \[\frac{ dx }{ dy } + P(y)x = Q(y)\]

Answer 26

Oh right, my equation is wrong

Answer 27

Try putting your equation in standard form first

Answer 28

You should see dx/dy, x on the left hand side and only y terms on the right hand side

Answer 29

\[\frac{ dx }{ dy } - y = 3-x\]

Answer 30

Hey its easy. take a deep breath and try again

Answer 31

You have :
\[\frac{ dx }{ dy } +x +\frac{ x }{ y }=3\]

Answer 32

Put it in your standard form

Answer 33

just factor out x from second and third terms

\[\frac{ dx }{ dy } +\color{red}{x +\frac{ x }{ y }}=3\]

Answer 34

huh, my mind is rambling

Answer 35

You know what
I already did that for you. Scroll up and look at my first reply in this thread

Answer 36

You will find the standard form in that reply

Answer 37

how it come y+1/y

Answer 38

Pull out x from the red terms and tell me what you get

\[\frac{ dx }{ dy } +\color{red}{x +\frac{ x }{ y }}=3\]

Answer 39

x(y+1)

Answer 40

you look distracted

Answer 41

oops. x(1+y/y)

Answer 42

\[\color{red}{x+\dfrac{x}{y}}\]

pulling out x gives you

\[\color{red}{x\left(1+\dfrac{1}{y}\right)}\]

Answer 43

yes

Answer 44

leave it like that maybe

Answer 45

\[\frac{ dx }{ dy } +\color{red}{x +\frac{ x }{ y }}=3\]


\[\frac{ dx }{ dy } +\color{red}{\left(1 +\frac{ 1 }{ y }\right)x}=3\]

Answer 46

Compare that with the standard form
\[\frac{ dx }{ dy } + P(y)x = Q(y)\]

Answer 47

P(y) = ?
Q(y) = ?

Answer 48

so p(y) = (1+1/y) and q(y) = 3

Answer 49

Yes

Answer 50

go ahead and find IF

Answer 51

I.F = \[\int\limits (1+\frac{ 1 }{ y })dy\]

Answer 52

\[\int\limits ydy+ lny\]

Answer 53

I.F = \[e^{\int\limits (1+\frac{ 1 }{ y })dy}\]

Answer 54

yes forgot the e

Answer 55

Do not show the integral and dy signs after evaluating the integral

Answer 56

\[\frac{ y^2 }{ 2 } + \ln y\]

Answer 57

sorry, i got y + lny

Answer 58

y +y^-1

Answer 59

Yes, so the IF is
\[e^{y+\ln y}\]

Answer 60

yes e^y = 1

Answer 61

yes

Answer 62

\[e^{y+\ln y} = e^y\bullet e^{\ln y} = e^y\bullet y = ye^y\]

Answer 63

so the solution now is
\[(ye^y)x = \int\limits 3 (ye^y)dy +c\]

Answer 64

Perfect!
you may try evaluating that integral too

Answer 65

\[\int\limits 3ye^y dy\]

Answer 66

https://www.wolframalpha.com/input/?i=Integrate%5B3ye%5Ey,%7By%7D%5D

Answer 67

\[\frac{ 3y^2e^y }{ 2}\]

Answer 68

is that right

Answer 69

try integration by parts int u dv = uv - int v du

Answer 70

u = 3y du = 3ydy dv = e^y v = ye^y

Answer 71

is that right

Answer 72

let u = x and dv = e^y dy

Answer 73

du = dx and v = ye^y

Answer 74

uv - int vdu

Answer 75

xye^y - int ye^y dx

Answer 76

then what next

Answer 77

oops, I meant let u = y and v = e^y dy (no x!)

Answer 78

so du =dy and v is ye^y right?

Answer 79

dv= e^y dy
v = e^y

Answer 80

oops sorry

Answer 81

so ye^y -int e^y dy

Answer 82

forgot the next process tho

Answer 83

can u do it

Answer 84

\[ \int e^y \ dy \]

Answer 85

is e^y

Answer 86

yes then

Answer 87

and put in the 3 from out front

Answer 88

and the constant of integration

Answer 89

so int 3e^y

Answer 90

\[ 3\int y \ e^y \ dy = 3( y e^y - e^y) + C\]

Answer 91

yes , but the anwer in book should be \[xy = 3(y-1) + ce ^{-y}\]

Answer 92

remember, you are doing
\[ (ye^y)x = \int\limits 3 (ye^y)dy +c \]

Answer 93

yes, so i will rewrite as
(ye6y)x = 3(ye^y-e^y)

Answer 94

how about
\[ ye^y \ x = 3(y-1)e^y + C \]
now multiply both sides by \(e^{-y} \)

Answer 95

yes, factor it out then i got xy = 3(y-1)+ ce^-y

Answer 96

thankyouuu