## kirbykirby 3 years ago x^2*y' + 3xy = 1 (Linear DE)

1. kirbykirby

I divide by x^2 to get y' + (3/x)y=1/x^2 My integration factor though gives me |x|^3

2. kirbykirby

I dunno what to do with this absolute value :S

3. shamim

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

4. kirbykirby

yes

5. anonymous

have you tried Cautiy Eiller..?

6. kirbykirby

WHat's Cautiy Eiller?

7. anonymous

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).

8. kirbykirby

(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|)

9. anonymous

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

10. kirbykirby

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

11. kirbykirby

after multiplying te integration factor

12. anonymous

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

13. anonymous

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

14. kirbykirby

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

15. anonymous

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

16. kirbykirby

How do you know if the x is always positive?

17. anonymous

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

18. kirbykirby

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?

19. anonymous

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

20. kirbykirby

No I was mainly concerned about the absolute value

21. kirbykirby

Thanks though !

22. shubhamsrg

Ignore the absolute value.

23. UnkleRhaukus

yeah.

24. anonymous

Yup