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kirbykirby
x^2*y' + 3xy = 1 (Linear DE)
I divide by x^2 to get y' + (3/x)y=1/x^2 My integration factor though gives me |x|^3
I dunno what to do with this absolute value :S
is it \[x ^{2}y \prime +3xy=1\]
have you tried Cautiy Eiller..?
WHat's Cautiy Eiller?
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).
(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|)
mm no.. sorry that won't work here..
I get \[|x|^3y' +\frac{3|x|^3}{x}=\frac{|x|^3}{x^2}\]
after multiplying te integration factor
the final step is then d/dx[x^3y]=int(x) which is simple.
excuse me it is x^3y=int(x) which is to say d/dx[x^3y]=x^3/x^2
But we hae absolute values no? Integral (1/x) = ln|x|
this yields x^3y=x^2/2 then jus divide by x^3 to get y=1/2(x^-1)
How do you know if the x is always positive?
Try reading some outside resources like http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx !!! I liked this site when i was a rook!
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?
otherwise though this solution looks correct..any problems you see?
No I was mainly concerned about the absolute value
Ignore the absolute value.