Math 310 Q3 Sol - MATH 310, Fall 2011: Differential...

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MATH 310, Fall 2011: Differential Equations Simon Fraser University Quiz 3: Solutions 1. Solve the initial value problem ty 0 = 2 y - 2 t 4 e - 2 t , t > 0; y (1) = y 0 . For which value of y 0 does the solution remain bounded as t → ∞ ? [Note: in finding the general solution you will need to integrate by parts once.] Solution: This is a first-order linear equation, so the first step in solving it is to put the DE in standard form: ty 0 - 2 y = - 2 t 4 e - 2 t = y 0 - 2 t y = - 2 t 3 e - 2 t ( t > 0) . The integrating factor is μ = exp[ Z p ( t ) dt ] = exp ± - Z 2 t dt ² = exp[ - 2 ln t ] = t - 2 ; multiplying the DE by the integrating factor, we have t - 2 y 0 - 2 t - 3 y = - 2 te - 2 t = ( t - 2 y ) 0 = - 2 te - 2 t , where in obtaining the last equation we identified the left-hand side as the total derivative of the product of the integrating factor and y . Now we can integrate both sides with respect to t , and integrate by parts: t - 2 y ( t ) = - 2 Z te - 2 t dt + c = - 2 ± - 1 2 te - 2 t - ³ - 1 2 ´Z e - 2 t dt ² + c = - 2
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Math 310 Q3 Sol - MATH 310, Fall 2011: Differential...

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