Question

solve the following differential eqn: x'-3x+x^2=0, if x=1 when t=0, determine the value of x when t=1

Answer 1

this one is separable

Answer 2

its not non-homo? i mean if i take x^2 to RHS ... then solve LHS as a compl soln... then RHS as particular soln???

Answer 3

perhaps you are right, sorry. one sec please...

Answer 4

this is linear, and so can be with an integrating factor I believe

Answer 5

no I am confused by the x I think :P
hm... bernoulli I guess
I'm not so good at those

Answer 6

not good?

Answer 7

I am not good at bernoulli problems, but others are

Answer 8

it is bernoulli's

Answer 9

thought so. well I'll leave it to you then :)

Answer 10

wats bernoulli?

Answer 11

http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

Answer 12

the trick with bernoulli's equation \[x \prime + p(t) x=q(t)x^n\] is to divide everything by x^n

Answer 13

wow thanks @TuringTest

@sirm3d my x^n is my x^2?

Answer 14

yes, you're right

Answer 15

x'/x^2 -3x/x^2=-1

Answer 16

x'/x^2-3/x=-1

Answer 17

or \[x^{-2}x \prime -3x^{-1}=-1\] where you sub \[v=x^{-1} \] . this should get you going.

Answer 18

i'm confused... v^(-1)x'-3v=-1?

Answer 19

whenever a sub is made in a DE, the derivative is generally expected of the sub. here \[v=x^{-1}\] so \[v \prime = -x^{-2} x \prime\]

Answer 20

can you do it from here?

Answer 21

:(

Answer 22

let me show it then
\[x^{-2}x \prime-3x^{-1}=-1\] by sub \[v=x^{-1}\] and \[v \prime = -x^{-2}x \prime\] we get \[-v \prime-3v=-1\]

Answer 23

@GeekChic_ are you still with me in this problem?

Answer 24

hi

Answer 25

@sirm3d u ther?

Answer 26

now that i got to v'-3v=-1 i solve it normally? complementary n particular solns?

Answer 27

Yc=Ae^(3v) and Yp=1/2

gen soln Y=Ae^(3v)+1/2

Answer 28

wait, a kid is messing with my mouse.

Answer 29

you can use an integrating factor. the integrating factor for \[y\prime+p(x) y=q(x)\] is \[v=\int\limits_{}^{}p(x)dx\] and the solution to the DE is \[vy=\int\limits_{}^{}vq(x)dx\]

Answer 30

let me correct the integrating factor. it should be \[v=e^{\int\limits_{}^{}p(x)dx}\] do you still remember this method of solving DE, @GeekChic_ ?

Answer 31

noooooo

Answer 32

oh no. one of the classes of 1st order linear DE is \[y\prime + p(x)y=q(x)\] and its solution is by using an integrating factor that will make the DE exact. the integrating factor turns out to be \[e^{\int\limits_{}^{}p(x)dx}\]

Answer 33

@GeekChic_ BUT you can use \[v_c =Ae^{-3x}\] and \[v_p = \frac{ 1 }{ 3 }\]

Answer 34

sorry it's not x but t as the independent variable.
the general solution is \[v=Ae^{-3t} + \frac{ 1 }{ 3}\] or in terms of x, \[x^{-1}=Ae^{-3t}+\frac{1}{3}\]
sub t=0, x=1 to find A, then
find x when t=1.

Answer 35

just one thing. the particular solution Yp may be difficult to find for the general function q(x) on the DE \[y\prime + p(x)y = q(x)\], so it is advisable to learn the method of integrating factor posted above.

Answer 36

wow... u are soooo good!!!! thank u sooo much

Answer 37

yw. =)

Answer 38

hi how are you?

Answer 39

@ sirmed, can do the beginning again plz

Answer 40

which part ^ is not clear to you?

Answer 41

ok never mind lo0l.. i just did it now and its ok now lol..thnx anyway.. :D

Answer 42

at first i did like this
x'-3x= -x^2 will that also work?

Answer 43

that's the first step, all right.

Answer 44

i didnt divide it to x^2
but instead directly work into sub x=u^-1, x'=-u^-2, will x^2=u^-2?

Answer 45

that's how bernoulli's equation is solved, the substitution \[u=x^{1-n}\]

Answer 46

ok i know that
but how about like this?
x'-3x= -x^2, x=u^-1, x'=-u^-2, will x^2=u^-2?
-u^-2 -3u^-1 = - u^-2 ?? will this work on the right side?

Answer 47

-u^-2 u'-3u^-1 = - u^-2 ?? will this work on the right side?

Answer 48

same result. after all, if u=x^-1, then x=u^-1

Answer 49

ok if continuing it would be
-u^-2 u'-3u^-1 = - u^-2
-----------------------
-u^-2

u'-3u=1 ? and your solution is

-u'-3u=-1 its the negative signs?

Answer 50

each term of your equation has a negative sign, so all terms are positive after division by -u^-2

Answer 51

ooops its +3u..lol :D sorry i didnt see that hahaha

Answer 52

ok then thank u very much ... lol :D

Answer 53

therefore we ended with the same equation. YW.

Answer 54

sometimes for just a little bit thing, everything messede up..lol... :D
thnx a lot and have fun solving now...