Use Improved Euler Solution y_{n+1} = y_{n} + (h/2)*( f(x_{n}, y_{n} ) + f( x_{n

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y_{n+1} = y_{n} + (h/2)*( f(x_{n}, y_{n} ) + f( x_{n} +h, y_{n} + f(x_{n}, y_{n} ) ) ) OR TAke k1= h f(x_{n}, y_{n} ) k2 = f(x_{n} + h, y_{n} +k1) y_{n+1} = y_{n} + (1/2)*( k1+k2) Starting with the initial condition x0; y0 and using this formula to successively find the values yk is known as the improved Euler’s Method. ///////////////////// Use Improved Euler’s method with a step size of 0.1 to approximate y(0.2), where y solves y'=-y+2x, y(0)=1 Improved Euler’s method 0.1 to approximate y(0.2), where y solves y'=-y+2x, y(0)=1 If n=0===> x0=0 AND y0=1 ==> f(x,y) = 2x-y k1= 0.1 ( 2x0 -y0) = 0.1*(0-1) = -0.1 k2= 0.1( 2(x0+h)-(y0+ k1) ) = 0.1( 2( 0+0.1)-(1-0.1) ) = -0.07 y(x0+h) = y(0+0.1)=y(x1) = y(0.1) = y0 + (k1+k2)/2 = 1 + ( -0.1-0.07) = 0.915 y(0.1) = 0.915 If n=1===> x1= x0+ h= 0+0.1=0.1 k1= 0.1 ( 2x1 -y1) = 0.1*( 2*(0.1)- 0.915) = -0.0715 k2= 0.1( 2(x1+h)-(y1+ k1) ) = 0.1( 2 ( 0.1+0.1)-( 0.915 -0.0715 ) ) = -0.04435 y(x1+h) = y(0.1+0.1)=y(x2) = y(0.2) = y1 + (k1+k2)/2 = y(0.2) = 0.915 + ( -0.0715 -0.04435) /2 ) = 0.857075

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