Q4B-230

# Q4B-230 - r 2(3-α r-2 α-1 = r 2 r α-1 So the general...

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Math 415 Name: Quiz 4B Verify that y 1 ( t ) = 1 and y 2 ( t ) = t 1 / 2 are solutions of the diﬀerential equation yy 00 + ( y 0 ) 2 = 0 for t > 0. Show that C 1 + C 2 t 1 / 2 is not a solution for this equation for all C 1 C 2 . y 1 y 00 1 + ( y 0 1 ) 2 = 1 · 0 - (0) 2 = 0 and y 2 y 00 2 + ( y 0 2 ) 2 = t 1 / 2 ( - 1 4 t - 3 / 2 ) + ( 1 2 t - 1 / 2 ) 2 = 0, so both y 1 and y 2 are solutions to the given equation. But if we let y 3 ( t ) = C 1 + C 2 t 1 / 2 then y 3 y 00 3 + ( y 0 3 ) 2 = ( C 1 + C 2 t 1 / 2 )( - 1 4 C 2 t - 3 / 2 ) + ( 1 2 C 2 t - 1 / 2 ) 2 = - C 1 C 2 4 t - 3 / 2 + 1 4 ( - C 2 2 + C 2 ) t - 1 , which is not equal to zero for all t unless C 2 = 0, or C 2 = 1 and C 1 = 0. Determine the value of α , if any, for which all solutions of y 00 + (3 - α ) y 0 - 2( α - 1) y = 0 tend to zero as t → ∞ . For which values of α , if any, do solutions become unbounded (positively or negatively) as t → ∞ . The characteristic equation for the given diﬀerential equation is 0 =
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Unformatted text preview: r 2 +(3-α ) r-2( α-1) = ( r +2)( r-( α-1)). So the general solution is y ( t ) = C 1 e-2 t + C 2 e ( α-1) t . The e-2 t term goes to zero as t approaches inﬁnity, so we only need to concern ourselves with the second term. If α-1 < 0 (i.e. α < 1), then the solution will approach zero as t → ∞ . If α > 1 then the solution will become unbounded as t → ∞ (whether to positive or negative inﬁnity depends on C 2 ). 1...
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