Q2B_130

# Q2B_130 - y t =-t 2 4 which is another solution of the...

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Math 415 Name: Quiz 2B Determine (without solving ) an interval in which the solution of ( t - 4) y 0 + (ln t ) y = 2 t, y (1) = 3 is certain to exist. Writing this linear equation in standard form we have y 0 + ln t t - 4 y = t t - 4 . The function t/ ( t - 4) is discontinuous at t = 4 and the function ln t/ ( t - 4) is continuous on the intervals (0 , 4) and (4 , ). Since our initial point t 0 = 1 is in the ﬁrst of these, this is where our solution exists. Show that y ( t ) = ct + c 2 is a solution of y 0 = - t + ( t 2 + 4 y ) 1 / 2 2 , y (2) = - 1 for any constant c . Show that there is no choice for the constant c that gives ˜
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Unformatted text preview: y ( t ) =-t 2 / 4, which is another solution of the diﬀerential equation. If y ( t ) = ct + c 2 then y ( t ) = c . On the other hand-t + ( t 2 + 4 y ( t )) 1 / 2 2 =-t + ( t 2 + 4 ct + 4 c 2 ) 1 / 2 2 =-t + ( t + 2 c ) 2 = c. So, no matter what c is equal to, y ( t ) is a solution. It should be clear that there is no c such that ct + c 2 =-t 2 / 4 since this would mean that, for all t , 1 4 t 2 + ct + c 2 = 0, which can’t happen. 1...
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## This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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