Question

Differential equations First order lnear

Answer 1

n is constant

Answer 2

looks like bernouli

Answer 3

you think so? Hmm

Answer 4

any idea how to solve this? =)

Answer 5

n is a constant means
it could be 2 or 3 or some other number... it doesn't change as x or y change..

Answer 6

its telling us not to bother about "n"

Answer 7

Ohh. Thank you so much.

Answer 8

can you finish off the rest ?

Answer 9

Wait can i just multiply the whole equation with 1/nx?

Answer 10

which equation ?

Answer 11

so that i'll arrive with dy/dx + 2y/nx = xy^n=1 / nx ?

Answer 12

the thing i posted. hehe

Answer 13

it'll look like the gen formula for first order linear

Answer 14

\[\dfrac{y'}{y^{n+1}}+ \dfrac{2}{nx}\frac{1}{y^n} = \frac{1}{n} \]

this is our equation, right ?

Answer 15

yes yes

Answer 16

but the orig eq

Answer 17

what i mean is that can i just multiply the orig equation with 1/nx?

Answer 18

yes, what good is that ?

Answer 19

\[nxy'+ 2y = xy^{n+1}\]

multiplying throguh out by 1/nx you get

\[y'+ \dfrac{2}{nx}y = \frac{1}{n}y^{n+1}\]

Answer 20

this is NOT a linear equation because the right hand side has \(y\) terms

Answer 21

for a linear wquation, the right hand side needs to be function of \(x\) alone, right ?

Answer 22

yes =) thank you

Answer 23

so ill integrate now the 2/nx right?

Answer 24

so that ill have the integrating factor

Answer 25

it wont work

Answer 26

Again, for integrating factor method to work, the right hand side needs to be a function of \(x\) alone

Answer 27

\[y'+ \dfrac{2}{nx}y = \frac{1}{n}\color{Red}{y^{n+1}}\]

thats the bad part

Answer 28

yes youre right

Answer 29

A linear equation needs to be of below form :
\[y' + p(x) y = q(x)\]

Answer 30

NO \(y\) terms on right hand side ok

Answer 31

so you can't use integrating factor method for your equation yet..

Answer 32

try it, it simply wont work

Answer 33

thats where bernouli substitution is helpful

Answer 34

what would be the px?

Answer 35

\[y'+ \dfrac{2}{nx}y = \frac{1}{n}\color{Red}{y^{n+1}}\]

Again, this is NOT a lienar equation yet

Answer 36

integrating factor stuff wont work

Answer 37

yes hehe sorry

Answer 38

v= y ^ 1-n, right? in bern eq?

Answer 39

but we can make it linear, lets start by dividing \(y^{n+1}\) both sides

Answer 40

\[y'+ \dfrac{2}{nx}y = \frac{1}{n}\color{Red}{y^{n+1}} \]

dividing through out by \(y^{n+1}\) gives
\[\dfrac{y'}{y^{n+1}}+ \dfrac{2}{nx}\dfrac{1}{y^n} = \frac{1}{n}\]

Answer 41

Now substitute \(\large v = \frac{1}{y^n}\)
differentiating both sides with respect to x gives \(\large v' = -\dfrac{n}{y^{n+1}} y' \implies -\dfrac{v'}{n} = \dfrac{y'}{y^{n+1}}\)

the equation becomes

\[-\dfrac{v'}{n} + \dfrac{2}{nx}v = \frac{1}{n}\]

which is same as

\[v'+ \dfrac{-2}{x}v = -1\]

Answer 42

Now thats a linear equation and you can use ur favorite integrating factor method :)

Answer 43

hahaha thank you so much. px would be -2/x right?

Answer 44

Yep!

Answer 45

is it okay to use the formula of bernoulli?

Answer 46

what formula ?

Answer 47

it should be okay to use any formula if it makes sense to you.. but i prefer this dividing and substitution process more as it lets me see whats going on..

Answer 48

hi c;

Answer 49

do you mean you got
\(\color{Red}{v}\)x^-2 = 1/x +c

Answer 50

yes

Answer 51

plug back the value of \(v\) :
\[v = \frac{1}{y^n}\]

Answer 52

the final solution will be
\[\frac{1}{y^n}x^{-2} = \frac{1}{x}+ c\]

Answer 53

you may make it look better by multiplying x^2 through out

Answer 54

omg got it lalala haha thank you so much thank you

Answer 55

=) @ShadowLegendX

Answer 56

the final solution will be
\[\frac{1}{y^n} =x+ cx^2\]

Answer 57

@ganeshie8 is da honey of all them bunnies

He knows dat stuff like all his rabbit holes

Answer 58

Awesome