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kirbykirbyBest ResponseYou've already chosen the best response.0
I divide by x^2 to get y' + (3/x)y=1/x^2 My integration factor though gives me x^3
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
I dunno what to do with this absolute value :S
 one year ago

shamimBest ResponseYou've already chosen the best response.0
is it \[x ^{2}y \prime +3xy=1\]
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
have you tried Cautiy Eiller..?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
WHat's Cautiy Eiller?
 one year ago

chemENGINEERBest ResponseYou've already chosen the best response.1
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).
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
(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)
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
mm no.. sorry that won't work here..
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
I get \[x^3y' +\frac{3x^3}{x}=\frac{x^3}{x^2}\]
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
after multiplying te integration factor
 one year ago

chemENGINEERBest ResponseYou've already chosen the best response.1
the final step is then d/dx[x^3y]=int(x) which is simple.
 one year ago

chemENGINEERBest ResponseYou've already chosen the best response.1
excuse me it is x^3y=int(x) which is to say d/dx[x^3y]=x^3/x^2
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
But we hae absolute values no? Integral (1/x) = lnx
 one year ago

chemENGINEERBest ResponseYou've already chosen the best response.1
this yields x^3y=x^2/2 then jus divide by x^3 to get y=1/2(x^1)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
How do you know if the x is always positive?
 one year ago

chemENGINEERBest ResponseYou've already chosen the best response.1
Try reading some outside resources like http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx !!! I liked this site when i was a rook!
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
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?
 one year ago

chemENGINEERBest ResponseYou've already chosen the best response.1
otherwise though this solution looks correct..any problems you see?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.0
No I was mainly concerned about the absolute value
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
Ignore the absolute value.
 one year ago
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