Integrating factor for solving equation problem.

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SUMMARY

The forum discussion focuses on solving the differential equation dy/dt = (-y/t) + 2 using the integrating factor method. The user initially finds the homogeneous solution, resulting in y = -t, and attempts to derive the general solution, ultimately arriving at y(t) = -tln(t^2) - tc. However, discrepancies arise when comparing this solution to the textbook answer, y(t) = t + c/t. Key errors identified include misapplication of the integrating factor and confusion between differentiation and integration steps.

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yaho8888
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Homework Statement



Find the gerneral solution of the differential equation below:
dy/dt=(-y/t)+2

Homework Equations



none

The Attempt at a Solution





my solution by using integrating factor:
1.find the homogenous solution first
dy/y = -1/t dt
you get ln(y) = -ln(t) when integrating both side.
you get y = -t

2. find gerenal soultion:
u(t) = 1/y(homogenous)
so in this case u(t) =1/-t, b(t) = 2

now apply integrating method:
(u(t)y(t))' = u(t)b(t)
(y(t)/-t)' = 2/-t
taking integral of both side we get
y(t)/-t = -2ln(t)+c = -ln(t^2)+c
then Y(t) = -t(-ln(t^2)+c) = -tln(t^2)+-tc
we get y(t) = -tln(t^2)+-tc for general solution.

but the book answer is y(t) = t+c/t

so did I do anything wrong, I don't really find anything wrong myself.
 
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yaho8888 said:
1.find the homogenous solution first
dy/y = -1/t dt
you get ln(y) = -ln(t) when integrating both side.
you get y = -t

It is not ln(-t), it is -ln(t).

2. find gerenal soultion:
u(t) = 1/y(homogenous)
so in this case u(t) =1/-t, b(t) = 2

Shouldn't u(t) = 1/t just by reading off the differential equation?

(y(t)/-t)' = 2/-t
taking integral of both side we get
y(t)/-t = -2ln(t)+c = -ln(t^2)+c

Differentiate, not integrate.
 

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