ps3solutions[1]

# ps3solutions[1] - origin, in Mathematica, the instructions...

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APMA 2101 Problem Set #3 Sample Solution Set (due 24 February 2009) Chapter 2: Supplementary Problem (2 points) For x(y) , we have dx/dy + x = y , with integrating factor exp(y) . Integrating both sides and using integration by parts once on the right-hand side gives x(y)=c*exp(-y)+y-1 . Chapter 2: Valentine’s Day Problem (4 points) Given (2 x - ax /( x 2 + y 2 ) 1/2 )+ (2 y+a - ay /( x 2 + y 2 ) 1/2 ) dy/dx = 0 , we identify M(x,y) and N(x,y) as the quantities in parentheses and check that M y =N x , so the equation is exact. We then integrate M dx = x 2 - a/ 2 2 x /( x 2 + y 2 ) 1/2 dx = x 2 - a ( x 2 + y 2 ) 1/2 + f(y) and integrate N dy = y 2 + ay - a/ 2 2 y /( x 2 + y 2 ) 1/2 dy = y 2 + ay - a ( x 2 + y 2 ) 1/2 + g(x) . Finally, the solution y(x) is defined implicitly by x 2 + y 2 + ay - a ( x 2 + y 2 ) 1/2 + c = 0 . Taking c = 0, in order to have a curve that passes through the
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Unformatted text preview: origin, in Mathematica, the instructions to plot the curve for a = 4 are: ContourPlot[ x^2+4y+y^2-4Sqrt[x^2+y^2], {x,-8,-8}, {y,-12,5}, Contours->{0}, PlotPoints->200 ] <Shift-Return> In Mathematica, the instructions to solve the equation for y[x] for a = 4 are: DSolve[2 x - 4 x / Sqrt[x^2+y[x]^2] + (2 y[x] + 4 - 4y[x] / Sqrt[ x^2 + y[x]^2] )y'[x]==0, y[x], x ] <Shift-Return> The trick, then, is to see if you can interpret the pages of results. Sometimes implicit definition is easier than an explicit solution, such as the one Mathematica is capable of providing. Section 2.4: Exact ODEs (4 points each)...
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## This note was uploaded on 11/02/2009 for the course APMA 2102 taught by Professor Keyes during the Spring '08 term at Columbia.

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