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

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

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

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

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

chemENGINEER
 one year ago
Best ResponseYou've already chosen the best response.1basically 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
 one year ago
Best 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)

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

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

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

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

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

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

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

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

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

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0COuld 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?

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

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

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