Question

Guys can you help me with this..Linear equation of order one, i try and cant do it

Answer 1

\[x(x^2+1)y'+2y= (x^2+1)^3\]

Answer 2

i will post what i done and correct me with my error ^.^

Answer 3

\[x(x^2+1)y'+2y=(x^2+1)^3\]\[y'+\frac{2}{x(x^2+1)}y=\frac{(x^2+1)^3}{x(x^2+1)}\]

Answer 4

form of a linear differential equation is \[y'+Py=Q\]\[\text{Integrating factor (I.F.)}=e^{\int{Pdx}}\]
final solution of the equation =\[y* \text{I.F.}= \int\limits Q* \text{I.F.} dx +k\]

Answer 5

\[p \int\limits_{}^{}\frac{ 2 }{ x(x^2+1) }dx\]

Answer 6

In ur question \[Q= \frac{(x^2+1)^3}{x(x^2+1)}\]\[\Large \text{I.F.}=e^{\int {\frac{2}{x(x^2+1)}dx}}\]

Answer 7

how can i integrate that :)

Answer 8

i will distribute the denominator?
\[\int\limits_{}^{} \frac{ 2}{ x^3+x }dx\]

Answer 9

\[2((x^-2/-2)+(lnx))\]

Answer 10

\[e^{-x^-2}+x+c\]

Answer 11

i end up
\[x e^{-x^-2}\]

Answer 12

\[\int\limits_{}^{}Q(x).I.F\]

Answer 13

\[yxe ^{-x^-2}=\int\limits_{}^{}\frac{ (x^2+1)^2 }{ x }\]

Answer 14

\[\int\limits \frac{2}{x(x^2+1)}dx\]\[ \int\limits \frac{2dx}{x^3(1+\frac{1}{x^2})}\]now substitute \(\Large \frac{1}{x^2}+1=u\) then \(\Large \frac{-2}{x^3}dx=du\)\[\int\limits \frac{-du}{u}\]replacing u again\[-\ln|u|\]\[-\ln \left( \frac{1}{x^2}+1 \right)\]

Answer 15

\[\Large \text{I.F.}=e^{\int \frac{2}{x(x^2+1)}dx}=e^{-\ln \left(1+\frac{1}{x^2}\right)}\]

Answer 16

(1/x^2 +1)^-1

Answer 17

yeah it is the integrating factor

Answer 18

can u help me with the right side again ^.^

Answer 19

y(1/x^2 +1)^-1=x + 1/x +c

Answer 20

\[\int\limits_{}^{}\frac{ (x^2+1)^2 }{ x }(1/x^2 +1)^-1\]

Answer 21

\[\int\limits_{}^{}\frac{ (x^2+1) }{ x }\]

Answer 22

@abb0t

Answer 23

\[ \frac{ y }{ (x^2+1)}=\frac{ (x+1)^2 }{ 2} +C\]

Answer 24

still end up the wrong answer

Answer 25

answer must be \[x^2y=\frac{ 1 }{ 4 }(x^2+1)^3+c(x^2+1)\]

Answer 26

Not really sure what's going on, but most people forget to slide their constant in the right place in the end, what I mean is you have something like this:

\[(\mu y)' = \mu r(x)\]
then
\[\mu y = \int \mu r(x) dx\]
then they'll usually write
\[y = \frac{1}{\mu} \int \mu r(x) dx\]
but they'll forget that when they integrate, their constant comes up here like this:

\[y = \frac{1}{\mu} \int \mu r(x) dx + C\]

when really it should be here like this since it distributes to both terms.

\[y = \frac{1}{\mu} \int \mu r(x) dx + \frac{C}{\mu}\]

Dunno if that helps or not, just a common mistake

Answer 27

anyone can point my mistake here ^.^ ... yeah @Kainui but i can see that error here..

Answer 28

cant*

Answer 29

Dunno, just start over from scratch maybe, it's good for ya. I'm working it out from start to finish with an integrating factor right now and I'll post the results in a sec.

Answer 30

yup @jiteshmeghwal9 , i continued it and i got an error, i dont know hwere it is. lol

Answer 31

\[\int \frac{(x^2+1)^3}{x(x^2+1)}*(1+\frac{1}{x^2})^{-1}+k\]

Answer 32

\[\int \frac{(x^2+1)^3}{x(x^2+1)}*\frac{x^2}{x^2+1} +k\]

Answer 33

\[\int\limits_{}^{}\frac{ (x^2+1) }{ x }\]

Answer 34

+k

Answer 35

\[\int x(x^2+1)dx +k\]

Answer 36

Solve this and u will get RHS

Answer 37

\[x(x^2+1)y'+2y=(x^2+1)^3\]
Multiply by an integrating factor, it's completely arbitrary, whatever helps to solve your problem.
\[\mu x(x^2+1)y'+2\mu y=\mu (x^2+1)^3\]
Now let's go ahead and make up a differential equation to define our integrating function,

\[[\mu x(x^2+1)]' = 2\mu\]

The point being just to undo a power rule, go ahead and work out the power rule from my backwards step here, this is the whole point.

\[(\mu x(x^2+1)y)' = \mu (x^2+1)^3\]

Now we can integrate, although we still have to solve for \(\mu\) to finish.

\[\mu x(x^2+1)y = \int \mu (x^2+1)^3dx\]
\[y = \frac{1}{\mu x(x^2+1)} \int \mu (x^2+1)^3dx\]

Alright time to solve our made up equation now.

\[[\mu x(x^2+1)]' = 2\mu\]
\[\mu' x(x^2+1) + \mu [x(x^2+1)]' = 2\mu\]
\[\frac{\mu'}{\mu} =\frac{2 - [x(x^2+1)]'}{x(x^2+1)} \]
\[\int \frac{d\mu}{\mu} = \int \frac{2 dx}{x(x^2+1)} -\int \frac{[x(x^2+1)]'}{x(x^2+1)}dx \]
\[\ln \mu = \int \frac{2 dx}{x(x^2+1)} - \ln[x(x^2+1)]\]

I don't know what that middle integral will be off the top of my head so I'll just stop here, the rest should follow though pretty easily since it's already all laid out. GL

Answer 38

haha I don't know if that helps or not to be honest, just did as much as was fun to me. :P

Answer 39

That is a massive work lol

Answer 40

@jiteshmeghwal9 do you have time to help me again? lol

Answer 41

\[(1+\frac{1}{x^2})^{-1}\]\[(\frac{x^2+1}{x^2})^{-1}\]\[(\frac{x^2}{x^2+1})\]

Answer 42

Got it ?

Answer 43

yip, thanks, moving on to next problem ^.^