H.W. 9.5 Solutions - loupe(anl2275 u2013 H.W 9.5 u2013...

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loupe (anl2275) – H.W. 9.5 – wolesensky – (53050) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points If y 0 is the particular solution of the differ- ential equation dy dx + 3 y + 2 e 5 x = 0 such that y (0) = 8, find the value of y 0 (1). 1. y 0 (1) = 1 4 e 5 + 33 4 e 3 2. y 0 (1) = - 1 4 e 5 - 33 4 e 3 3. y 0 (1) = - 1 4 e 5 + 33 4 e 3 correct 4. y 0 (1) = - 1 4 e 5 - 31 4 e 3 5. y 0 (1) = 1 4 e 5 + 31 4 e 3 Explanation: For the first order differential equation ( ) dy dx + 3 y + 2 e 5 x = 0 the integrating factor is μ = e integraltext 3 dx = e 3 x . After multiplying both sides of ( ) by e 3 x we can thus rewrite the equation as d dx parenleftBig y e 3 x parenrightBig = - 2 e 8 x . Consequently, the general solution of ( ) is given by y e 3 x = - 2 integraldisplay e 8 x dx = - 1 4 e 8 x + C where C is an arbitrary constant. For the particular solution y 0 the value of C is deter- mined by the condition y (0) = 8 since y (0) = 8 = 8 = - 1 4 + C. Thus y 0 = - 1 4 e 5 x + 33 4 e 3 x . Hence y 0 (1) = - 1 4 e 5 + 33 4 e 3 . 002 10.0points If y 1

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