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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0have you tried Cautiy Eiller..?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0basically 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
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mm no.. sorry that won't work here..

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the final step is then d/dx[x^3y]=int(x) which is simple.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0excuse me it is x^3y=int(x) which is to say d/dx[x^3y]=x^3/x^2

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this yields x^3y=x^2/2 then jus divide by x^3 to get y=1/2(x^1)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Try reading some outside resources like http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx !!! I liked this site when i was a rook!

kirbykirby
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0otherwise though this solution looks correct..any problems you see?

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

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