Question

Find the general solution of y'+(1/x-1)y=-2/x.

Answer 1

@iambatman Knows how to do this, I don't Sorry :C

Answer 2

@iambatman

Answer 3

Differential equations? What have you done so far?

Answer 4

Yes, I tried to do integration by parts but it only complicated the problem more.

Answer 5

Please help me. I've stucked on this problem for days.

Answer 6

\[y'+\frac{ 1 }{ x-1 }y=-\frac{ 2 }{ x }\] right?

Answer 7

No, the middle part is ((1/x)-1)y

Answer 8

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

Answer 9

Yes, please continue.

Answer 10

He is not going to give you the answer, he is going to help you and teach you how to get the answer yourself ♩ヽ(・ω・ヽ*)♬

Answer 11

Okay.

Answer 12

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

solving as a linear equation right?

Answer 13

Yes.

Answer 14

I was about to call ganeshie as back up but he's already here :3

Answer 15

hey are you the same Idealist that kept me waiting the other day :o

Answer 16

Yes, the other website doesn't work so I've created a new account.

Answer 17

\[\huge u = e ^{\int\limits \frac{ -x+1 }{ x }dx}=e ^{-x}x?\] so

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

multiplying both sides by u, Idk if this is right though, ganeshie knows more...

Answer 18

Okay haha! i think iambatman is already helping you, ima just watch how you work this... :) this is really an interesting problem if you're doing linear DEs for the first time !

Answer 19

Nuuu go ahead ganeshie :3

Answer 20

So many smart people (づ｡◕‿‿◕｡)づ

Answer 21

that looks perfect to me !

Answer 22

the whole idea of multiplying \(\large u\) both sides is to write the left side as \(\large (uy)'\) , @Idealist10 do you see how to apply the product rule in reverse on left side ?

Answer 23

So the integrating factor is (e^-x)(x)?

Answer 24

Substitute \[e ^{-x}(1-x) = \frac{ d }{ dx }\left( \frac{ x }{ e^x } \right)\]

Answer 25

In the product rule (uy)', since we already found u, the integrating factor, what's y?

Answer 26

Mhm, Well you have \[\frac{ x*\frac{ dy }{ dx } }{ e^x }+\frac{ d }{ dx }\left( \frac{ x }{ e^x } \right)y=-\frac{ 2 }{ e^x }\]

So now you want to apply reverse product rule,

\[\frac{ d }{ dx }\left( \frac{ xy }{ e^x } \right)=-\frac{ 2 }{ e^x }\]

Now you can integrate...

\[\int\limits \frac{ d }{ dx }\left( \frac{ xy }{ e^x } \right)dx=\int\limits \left( -\frac{ 2 }{ e^x } \right)dx\]

the reverse product rule is \[g \frac{ df }{ dx }+f \frac{ dg }{ dx }=\frac{ d }{ dx }(fg)\]

Once again thought I'm not entirely sure if this is 100% correct haha.

Answer 27

thats looks good to me, @Idealist10 : after some practice, you can blindly write the left side as \(\large (\color{Red}{u}y)'\)...
to your earlier question, \(y\) is just \(y\), left side becomes \(\large (\color{red}{\dfrac{x}{e^x}}y)'\)

(shown as dy/dx notation in above reply)

Answer 28

Yeah sorry lost connection

Answer 29

Does this make sense?

Answer 30

Thank you so much for the help. I'm sorry it took long to understand. I just find this topic hard.

Answer 31

No worries, no reason to apologize for that :). I actually like it when people take their time and try to learn subjects ^.^