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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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We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
plz help kr do koi tu
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
has a unique solution y(t) on some t-interval containing t0.
i hate this subject..
jab ans ata he na ho tu hate he karna ha
but i love this subject
nafrat ni karna es subject sy. fail hojaye ga meri tarah