9.10HW-solutions

9.10HW-solutions - .2714 10 2.2.5385.4908 9.7 3.3.7248.6651...

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MATH 2930: HW Solutions: Assigned September 10, 2010 Prepared by Alex Moore (problem #22 graded) Section 2.6 22.) ( x + 2) sin y d x + x cos y d y = 0 ∂M ∂y = (( x + 2) sin y ) ∂y = ( x + 2) cos y ∂N ∂x = ( x cos y ) ∂x = cos y ∂M ∂y 6 = ∂N ∂x This is not exact. We multiply both sides by the integrating factor μ = xe x and recalculate the partial derivatives: ( x + 2) xe x sin y d x + x 2 e x cos y d y = 0 ∂M ∂y = (( x + 2) xe x sin y ) ∂y = ( x + 2) xe x cos y ∂N ∂x = ( x 2 e x cos y ) ∂x = ( x + 2) xe x cos y ∂M ∂y = ∂N ∂x Now the equation is exact. To have an easier integral, I’m going to integrate N d y instead of M : ψ = Z N d y = Z ( x 2 e x cos y )d x = x 2 e x sin y + g ( x ) 1

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ψ x = ∂ψ ∂x = x 2 e x sin y + 2 xe x sin y + g 0 ( x ) Comparing this to M d y , we see that g 0 ( x ) = 0 and g is a constant. The final answer becomes: x 2 e x sin y = C Section 2.7 4a. and d.) Euler’s method is given by the following equation: y n +1 = y n + f ( t n , y n ) h For our specific problem, this turns into: y 0 = 0 y 1 = y 0 + f (0 , 0) h = 0 + (3)( . 1) = . 3 y 2 = y 1 + f ( . 1 , . 3) h = . 3+ = . 3 . . . . . . y n +1 = y n + ( - 2 y n + 3 cos t n ) h The final table is given below: n t y n Exact % Error 0 0 0 0 0% 1
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Unformatted text preview: .2714 10% 2 .2 .5385 .4908 9.7% 3 .3 .7248 .6651 8.9 % 4 .4 .8665 .7997 8.3% 2 The exact numbers are from the solution to the diﬀerential equation. The equation can be put into the form of a ﬁrst order linear equation: y + 2 y = 3 cos t with p ( t ) = 2. Thus our integrating factor is μ = e R 2d t = e 2 t . Multiplying both sides of our diﬀerential equation and simplifying gives: y e 2 t + 2 ye 2 t = 3 e 2 t cos t d d t [ e 2 t y ] = 3 e 2 t cos t e 2 t y = Z 3 e 2 t cos t d t = 3 5 e 2 t (sin t + 2 cos t ) + C ⇒ y ( t ) = 3 5 (sin t + 2 cos t ) + C e 2 t Plugging in the initial condition y (0) = 0 gives C =-6 5 . The ﬁnal answer is: y ( t ) = 3 5 (sin t + 2 cos t ) +-6 5 e 2 t 3...
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