chapter13ProblemsAndSolutions

# chapter13ProblemsAndSolutions - Chapter 13 More...

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Unformatted text preview: Chapter 13 More Differential Equations 13.1 Consider the differential equation dy dt = a- by where a , b are constants. (a) Show that the function y ( t ) = a b- Ce- bt satisfies the above differential equation for any constant C . (b) Show that by setting C = a b- y we also satisfy the initial condition y (0) = y . Remark: You have now shown that the function y ( t ) = parenleftBig y- a b parenrightBig e- bt + a b is a solution to the initial value problem (i.e differential equation plus initial condition) dy dt = a- by, y (0) = y . Detailed Solution: (a) Differentiating y ( t ) = a/b- Ce- bt leads to dy/dt = bCe- bt . On the other hand, a- by = a- b ( a b- Ce- bt ) = bCe- bt This verifies that the function satisfies the given differential equation. v.2005.1 - September 4, 2009 1 Math 102 Problems Chapter 13 (b) y (0) = a/b- Ce = a/b- C. Substituting in C = a/b- y : y (0) = a/b- ( a/b- y ) = y . This verifies that the function also satisfies the initial condition. 13.2 For each of the following, show the given function y is a solution to the given differential equation. (a) t · dy dt = 3 y , y = 2 t 3 . (b) d 2 y dt 2 + y = 0, y =- 2 sin t + 3 cos t . (c) d 2 y dt 2- 2 dy dt + y = 6 e t , y = 3 t 2 e t . Detailed Solution: (a) y = 2 t 3 , dy dt = 6 t 2 . Substituting dy dt into the differential equation gives t dy dt = t · 6 t 2 = 6 t 3 = 3 y . So y = 2 t 3 is a solution to t dy dt = 3 y . (b) y =- 2 sin t + 3 cos t dy dt =- 2 cos t- 3 sin t d 2 y dt 2 = 2 sin t- 3 cos t Substituting y and d 2 y dt 2 into parenleftbigg d 2 y dt 2 + y parenrightbigg gives (2 sin t- 3 cos t ) + (- 2 sin t + 3 cos t ) = 0. So y =- 2 sin t + 3 cos t is a solution to d 2 y dt 2 + y = 0. (c) y = 3 t 2 e t dy dt = 6 te t + 3 t 2 e t d 2 y dt 2 = 6 e t + 6 te t + 6 te t + 3 t 2 e t = 6 e t + 12 te t + 3 t 2 e t v.2005.1 - September 4, 2009 2 Math 102 Problems Chapter 13 Substituting y , dy dt and d 2 y dt 2 into parenleftbigg d 2 y dt 2- 2 dy dt + y parenrightbigg gives d 2 y dt 2- 2 dy dt + y = 6 e t + 12 te t + 3 t 2 e t- 2(6 te t + 3 t 2 e t ) + 3 t 2 e t = 6 e t y = 3 t 2 e t satisfies the differential equation d 2 y dt 2- 2 dy dt + y = 6 e t . 13.3 Show the function determined by the equation 2 x 2 + xy- y 2 = C , where C is a constant and 2 y negationslash = x , is a solution to the differential equation ( x- 2 y ) dy dx =- 4 x- y . Detailed Solution: To find dy dx , we apply the method of implicit differentiation to 2 x 2 + xy- y 2 = C . d dx ( 2 x 2 + xy- y 2 ) = d dx ( C ) 4 x + parenleftbigg y + x dy dx parenrightbigg- 2 y dy dx = 0 dy dx ( x- 2 y ) =- 4 x- y dy dx =- 4 x- y x- 2 y Substituting dy dx into ( x- 2 y ) dy dx gives ( x- 2 y )- 4 x- y x- 2 y =- 4 x- y . So the function given by 2 x 2 + xy- y 2 = C is a solution to the differential equation ( x- 2 y ) dy dx =- 4 x- y ....
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chapter13ProblemsAndSolutions - Chapter 13 More...

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