kirbykirby
  • kirbykirby
x^2*y' + 3xy = 1 (Linear DE)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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kirbykirby
  • kirbykirby
I divide by x^2 to get y' + (3/x)y=1/x^2 My integration factor though gives me |x|^3
kirbykirby
  • kirbykirby
I dunno what to do with this absolute value :S
shamim
  • shamim
is it \[x ^{2}y \prime +3xy=1\]

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More answers

kirbykirby
  • kirbykirby
yes
nubeer
  • nubeer
have you tried Cautiy Eiller..?
kirbykirby
  • kirbykirby
WHat's Cautiy Eiller?
anonymous
  • 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).
kirbykirby
  • 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|)
nubeer
  • nubeer
mm no.. sorry that won't work here..
kirbykirby
  • kirbykirby
I get \[|x|^3y' +\frac{3|x|^3}{x}=\frac{|x|^3}{x^2}\]
kirbykirby
  • kirbykirby
after multiplying te integration factor
anonymous
  • anonymous
the final step is then d/dx[x^3y]=int(x) which is simple.
anonymous
  • anonymous
excuse me it is x^3y=int(x) which is to say d/dx[x^3y]=x^3/x^2
kirbykirby
  • kirbykirby
But we hae absolute values no? Integral (1/x) = ln|x|
anonymous
  • anonymous
this yields x^3y=x^2/2 then jus divide by x^3 to get y=1/2(x^-1)
kirbykirby
  • kirbykirby
How do you know if the x is always positive?
anonymous
  • 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!
kirbykirby
  • 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?
anonymous
  • anonymous
otherwise though this solution looks correct..any problems you see?
kirbykirby
  • kirbykirby
No I was mainly concerned about the absolute value
kirbykirby
  • kirbykirby
Thanks though !
shubhamsrg
  • shubhamsrg
Ignore the absolute value.
UnkleRhaukus
  • UnkleRhaukus
yeah.
SWAG
  • SWAG
Yup

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