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Discussion Topic:
Under what conditions the solution of the first order ordinary differential equation exist?
Furthermore, without solving the following initial value problem determine the interval in which the solution is certain to exist.
Opening Date: August 13, 2014 at 12:01 AM
Closing Date: August 18, 2014 at 11:59 PM
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2nd question ka kia answer hy? kisi ko pata hy kia?
sanu pehla nhi araha ap dusra bol rahy kamal kerty hen pandy g...!
Find the general solution by using an integrating factor:
y' + ytant = sint
dy / dt + ytant = sint
dy / dt + P(t)y = f(t)
P(t) = tant
f(t) = sint
I(t) = ℮^[∫ P(t) dt]
I(t) = ℮^(∫ tant dt)
I(t) = ℮^(-ln|cost|)
I(t) = ℮^(ln|1 / cost|)
I(t) = 1 / cost
I(t)y = ∫ I(t)f(t) dt
y / cost = ∫ tant dt
y / cost = -ln|cost| + C
y / cost = C - ln|cost|
y = cost(C - ln|cost|)
Find the particular solution by solving for the constant:
When t = π, y = 0
-C = 0
C = 0
y = -costln|cost|
Suppose that f(t,y) and partial deriavative of f(t,y) are continuous on a closed rectangle R of the ty -plane. If , then the IVP
y'=f(t,y) ,y(t0)=y0
has a unique solution y(t) on some t-interval containing t0.
but i love this subject
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