nijhu
dw:1349516917936:dw



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hartnn
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draw it out again

nijhu
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ok

nijhu
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dw:1349517106533:dw

hartnn
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u need to evaluate that ?
its w.r.t du , right ?
which topic does this question belong ?

nijhu
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i need to solve this by using convollution theorem

nijhu
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it is a laplace transform

hartnn
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Laplace or Fourier ? F = Fourier generally

nijhu
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we r following th e book laplace transform by spiegel

nijhu
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and the topic is alpplications to differential equations by using inverse laplace

nijhu
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anyone got my answr

nijhu
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?

hartnn
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\(\large \frac{s}{5}\frac{5}{(s^2+9)(s^2+4)}= \quad \frac{s}{5}\frac{(s^2+9)(s^2+4)}{(s^2+9)(s^2+4)}\)
now just separate denominator.
got it ?

nijhu
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dw:1349520425747:dw

hartnn
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don't include s.
lets just do partial fraction for 1/(s^2+4)(s^2+9)
after that we multiply s to both terms

hartnn
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dw:1349520610397:dw

nijhu
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hmmm then?

hartnn
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what vales of A and b u got ?

nijhu
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A=1/85 and B=1/25

hartnn
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how !
1= A (s^2+4) + B (s^2+9)
put s^2 = 9
1= A (9+4)
1= A (5)
A = 1/5
got this ?

hartnn
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similarly find B

nijhu
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B=1/5

hartnn
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thats correct.
so now u got where that 1/5 came from ?

nijhu
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lol....

nijhu
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got it

nijhu
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u r very helpfull

nijhu
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thanks a lot man!

hartnn
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welcome ^_^

nijhu
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dw:1349522140539:dw

hartnn
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thats just separation of variables, which step did u not get ?

nijhu
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(s^2+4)dy/ds+sy=0....how did the minus sign cancel out?

nijhu
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dw:1349522675157:dw

nijhu
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dw:1349522813018:dw

hartnn
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u r rite, even i don't see, how they got that +
maybe printing mistake in book.....

nijhu
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my sir showed us a similar problem where the condition Y(0)=1,Y(pie)=0 and Y'(0)=c