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Consider the first order initial value problem y′=y+2x−x2,y(0)=1,(0≤x<∞) with exact solutions y(x)=x2+ex. For x =0.1 , the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta with step-size h = 0.1 is0.063

Answer» Consider the first order initial value problem y=y+2xx2,y(0)=1,(0x<) with exact solutions y(x)=x2+ex. For x =0.1 , the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta with step-size h = 0.1 is
  1. 0.063


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