Question

whats the general answer of (x^2)y'+3xy=sinx/x ?

Answer 1

What's the solution of

\[ x^2 y' + 3x y = 0 \]

Answer 2

I would think about the integrating factor. Then your problem becomes easy.

Answer 3

These questions are fun because they linear differential equations :)

Answer 4

isnt it seperably?

Answer 5

The equation James wrote is.

Answer 6

Integrating factor

Answer 7

But the equation you wrote is not.

Answer 8

Do you want me to show you the general way to solve these types if it is okay with James?

Answer 9

yes.

Answer 10

sure

Answer 11

[
You might also find this helpful:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/
]

Answer 12

Ok so we have something that can be written in this form:
\[y'+p(x)y=q(x)\]
Now if we multiply both sides by v>0 such that v'=vp since product rule is vy'+v'y=(vy)'
\[vy'+vpy=vq\]
So vp=v' so this means we have
\[vy'+v'y=vq => (vy)'=vq\]
Now we can integrate both sides:
\[vy=\int\limits_{}^{}vq dx +C\]
And solve for y
\[y=\frac{1}{v} \int\limits_{}^{}vq dx +C\]
But what is v?
Recall v'=vp We can solve this for v by using separation of variables
v'=dv/dx by the way and p is a function of x
\[\frac{dv}{dx}=vp => \frac{dv}{v}=p dx => \ln(v)=\int\limits_{}^{}p dx+c => v=e^{\int\limits_{}^{} p dx+c} \text{ note take c=0}\]
\[ => v=e^{\int\limits_{}^{}p dx} \text{ this is call the integrating factor} \]

Answer 13

mistake

Answer 14

\[y=\frac{1}{v} \int\limits\limits_{}^{}vq dx +\frac{C}{v}\]

Answer 15

i forgot to divide the constant by v above

Answer 16

for particular solution to this one, you might be to do variation of parameter

Answer 17

Do you have any questions
put into this form y'+py=q
and then find what we called the integrating factor
and if you want you can use that formula I have for finding y

Answer 18

im trying to solve it but im confused

Answer 19

it takes a little time for me ^.^

Answer 20

I understand work takes time

Answer 21

see im try but it seems im really dumb
first i divide both side by x^2 and get y' + 3y/x = sinx/x^3
second i integrating factor is e^(integral of p) (and p is equal to 3/x right?) and i get e^3lnx
equation now looks like y'(e^3lnx) + (e^3lnx)*3y/x = (e^3lnx)*sinx/x^3
what should i do with this 1. its bigger than its initial form /cry

Answer 22

or x^3 right?

Answer 23

e^(3lnx)=e^(lnx^3)=x^3

Answer 24

\[x^3y'+x^3 \cdot \frac{3}{x}y=\sin(x)\]
\[x^3y'+3x^2y=\sin(x)\]
remember product rule
\[(x^3y)'=\sin(x)\]

Answer 25

can you get it form there
just integrate both sides now

Answer 26

from*

Answer 27

lol sorry

Answer 28

its ok

Answer 29

so in the end answer is (c-cosx)/x^3 but im having trouble in typing in this form because my mainboard is in rma.(my current system is amd 3500+ /cry). thnx for help.

Answer 30

8 hours till TEST

Answer 31

lets check
I'm going to integrate the last equation I wrote.
\[x^3y=-\cos(x)+C\]
yes thats right! :)
gj

Answer 32

i don't know what all that other you are talking about is
sorry

Answer 33

some kind of computer talk i believe i guess