Initial Value Problem. I'm really confused just need some help

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SUMMARY

The forum discussion centers on solving the initial value problem defined by the differential equation (dx/dt) - xt = -t with the initial condition x(0) = 2. The user successfully separates variables and integrates to find ln[x-1] = (1/2)t^2. The main challenge arises in calculating x(1), where the solution involves substituting t = 1 into the equation x - 1 = e^(t^2/2). The final step is to compute the numerical value of x(1) by evaluating the expression.

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mazz1801
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Homework Statement



Solve the following Initial Value problem for x(t) and give the value of x(1)

Homework Equations



(dx/dt)-xt=-t , x(0)=2

The Attempt at a Solution



(dx/dt)-xt = -t
(dx/dt) = xt-t
(dx/dt) = t(x-1)
(1/(x-1)) (dx/dt) = t
(1/(x-1)) dx = t dt

Then I integrate both sides.

∫(1/(x-1)) dx = ∫t dt
ln[x-1] + C = (1/2)t^2

I put x=2 and t=0
ln[1] + C = (1/2)(0)^2
0 + C = 0
C = 0

So my problem is the x(1) bit... I don't know where to go with this... would I just make t=1 and solve for x and give a numerical answer by subbing i back in?
Or am I way of the mark and have everything wrong :O
 
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welcome to pf!

hi mazz1801! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
mazz1801 said:
ln[x-1] + C = (1/2)t^2

C = 0

So my problem is the x(1) bit... I don't know where to go with this... would I just make t=1 and solve for x and give a numerical answer by subbing i back in?

("subbing i"? :confused:)

yes, that's fine :smile:

you should now get rid of the ln by writing x-1 = et2/2

then put t = 1
 
Thank you very much tiny-tim :)
 

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