Question

x^2*y' + 3xy = 1 (Linear DE)

Answer 1

I divide by x^2 to get
y' + (3/x)y=1/x^2

My integration factor though gives me |x|^3

Answer 2

I dunno what to do with this absolute value :S

Answer 3

is it \[x ^{2}y \prime +3xy=1\]

Answer 4

yes

Answer 5

have you tried Cautiy Eiller..?

Answer 6

WHat's Cautiy Eiller?

Answer 7

basically
Put the differential equation in the correct initial form
Find the integrating factor
Multiply everything in the differential equation by and verify that the left side becomes the product rule and write it as such.
Integrate both sides, make sure you properly deal with the constant of integration.
Solve for the solution y(t).

Answer 8

(I tried seeing what wolfram alpha does, but they always do Integral (1/x) as ln x, but we are supposed to do it as ln |x|)

Answer 9

mm no.. sorry that won't work here..

Answer 10

I get \[|x|^3y' +\frac{3|x|^3}{x}=\frac{|x|^3}{x^2}\]

Answer 11

after multiplying te integration factor

Answer 12

the final step is then d/dx[x^3y]=int(x) which is simple.

Answer 13

excuse me it is x^3y=int(x) which is to say d/dx[x^3y]=x^3/x^2

Answer 14

But we hae absolute values no? Integral (1/x) = ln|x|

Answer 15

this yields x^3y=x^2/2 then jus divide by x^3 to get y=1/2(x^-1)

Answer 16

How do you know if the x is always positive?

Answer 17

Try reading some outside resources like http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx !!! I liked this site when i was a rook!

Answer 18

COuld it be because if we say x>0, then the drop the absolute value, but if x <0, then you put -x^3 everywhere (but since the whole equation is negative, you just "divide by -1" o_o?

Answer 19

otherwise though this solution looks correct..any problems you see?

Answer 20

No I was mainly concerned about the absolute value

Answer 21

Thanks though !

Answer 22

Ignore the absolute value.

Answer 23

yeah.

Answer 24

Yup