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Quiz 1 / Re: Q1 problem 2 (Night sections)
« on: September 26, 2014, 08:01:30 PM »
I dont use intergrating factor for this problem, instead I use an alternative approach
y' - 2y = e ^(2t) ---------(1)
--> z' -2z = 0 (assume the R.H.E. =0), then dz\z=2dt --> ∫dz/z =2dt --> ln |z| = 2t+C1
-->|z| = C1e^(2t) --> z = Ce^(2t), C = ±C1
-->now let C = C(t), such that y= Ce^(2t)--------------(2)
plug (2) into (1), we get C'e^(2t) + 2Ce^(2t)- 2Ce^(2t) = e^(2t) --->C'e^(2t) = e^(2t)
--->C' = 1 ---> C = t + C2 ---> y = Ce^(2t) = ( t + C2)e^(2t) ----(3)
now plug in (0,2) into (3) -->2 =C2 -->therefore y = ( t + 2)e^(2t)
y' - 2y = e ^(2t) ---------(1)
--> z' -2z = 0 (assume the R.H.E. =0), then dz\z=2dt --> ∫dz/z =2dt --> ln |z| = 2t+C1
-->|z| = C1e^(2t) --> z = Ce^(2t), C = ±C1
-->now let C = C(t), such that y= Ce^(2t)--------------(2)
plug (2) into (1), we get C'e^(2t) + 2Ce^(2t)- 2Ce^(2t) = e^(2t) --->C'e^(2t) = e^(2t)
--->C' = 1 ---> C = t + C2 ---> y = Ce^(2t) = ( t + C2)e^(2t) ----(3)
now plug in (0,2) into (3) -->2 =C2 -->therefore y = ( t + 2)e^(2t)