Question

What is the solution to the following bernoulli de?
\[t^2 dy/dx+y^2=ty\]
It looks like it is homogenous because if you divide through by \[t^2\] it will be \[y/t\] all over. Also if it's a bernoulli, wouldn't \[v=y\], which isn't very useful. Any thoughts?

Answer 1

here t is constant only x and y are variables
so it reduces to\[\frac{dy}{ty-y^2}=\frac1{t^2}dx\]just solve it taking t as constant

Answer 2

it looks like from \[t^2 dy/dx + y^2 =ty\] you can divide from both sides by \[t^2\]to get \[dy/dx + y^2/t^2 =y/t\]and then through some algebra you can rearrange it as\[ dy/dx -y/t =-y^2/t^2\]where then you can use a substitution and let \[w=-y^{-3}\]

Answer 3

then you can solve the differential equation throught the linear method using the substitution from w

Answer 4

from my equation \[\frac1t \int\limits \frac{y+(t-y)}{y(t-y)}dy=\frac1{t^2} \int\limits dx\]\[-\ln(y-t)+\ln(y)=\frac xt+c\]\[\ln \frac y{y-t}=\frac xt +c\]

Answer 5

its easy approach

Answer 6

is that using bernoulli DE?

Answer 7

no.

Answer 8

using bernouli's \[y^{-2}dy/dx+1/t^2=1/(ty)\]\[y^{-2}dy/dx-1/(ty)=-1/t^2\]take 1/y=z dz/dx=-1/y^2dy/dx
so \[-dz/dx-z/t=-1/t^2\]\[dz/dx+x/t=-1/t^2\]now its linear

Answer 9

use leibnitz' rule to find the solution

Answer 10

bleh I mean \[w = y^{-1}\]so then \[dw/dx = -y^{-2}dy/dx\]and rearranging gives\[dy/dx = -y^2dw/dx\]so then the substitued equation becomes\[-y^2dw/dx - y/t = -y^2/t^2\]simlifying further gives\[dw/dx + w/t = 1/t^2\]then you can use the integrating factor to let \[e ^{\int\limits_{ }^{ }(1/t) dt}\]which is \[e^{\ln t} = t\]and becomes
\[wt = \int\limits\limits_{ }^{ }1/t^2 dt = -t^{-1}+c\]therefore your equation becomes\[w=-1/t^2+c/t\]and subsituting back in for y you get\[y^{-1} = -1/t^2 +c/t\]as your solution for the differential equation using the bernoulli method

Answer 11

what u have done??

Answer 12

u have dw/dx not dw/dt..

Answer 13

o bleh i keep typing x for t

Answer 14

so how could be the IF t???

Answer 15

wait wait..

Answer 16

u have dw/dx+w/t=1/t^2 right??

Answer 17

yea

Answer 18

when you divide\[−y^2dw/dx−y/t=−y^2/t^2\]by\[−y^2\]from both sides you get \[dw/dx+y^{-1}/t=1/t^2\]but remember that we sub \[w=y^{-1}\]so it becomes\[dw/dx+w/t=1/t^2\]

Answer 19

dw/dx=(1-wt)/t^2
ok??

Answer 20

dw/(1-wt)=dx/t^2 ok???

Answer 21

here t is not a variable u need to do work with w and x only...

Answer 22

do u agree??

Answer 23

.... omg i read the problem wrong O.O

Answer 24

i thought it was \[t^2dy/dt+y^2=ty\] not \[t^2dy/dx+y2=ty\]

Answer 25

here u get \[\int\limits dw/(1-wt)=int dx/t^2\]\[\ln(1-wt)/(-t)=x/t^2+c\]\[\ln((y-t)/y)=-x-ct\]\[\ln(y/(y-t))=x+c'\]that matches with my previous work

Answer 26

so u agree with me now.

Answer 27

yea, made an error in what i was taking it wrt

Answer 28

My book says the answer is \[e ^{t/y}=ct\]...