Euler's Method

Linearization Linearization is a topic you may not have covered in Calculus 1. Here, L(x) is the tangent line to f (x) at x = x0 . (x1 , L(x1 )) L(x) f (x) (x1 , f (x1 )) (x0 , f (x0 )) f (x1 ) ≈ L(x1 ) = f (x0 ) + f 0 (x0 ) ∗ (x1 − x0 ) Linearization Suppose we have the first-order IVP dy = f (x, y ), y (a) = y0 dx and we want to approximate the value of the solution at x = b, i.e. approximate y (b). According to Linearization,   dy y (b) ≈ y (a) + (b − a) = y (a) + f (a, y (a)) · (b − a). dx (a,y (a)) This approximation gets worse as b gets farther away from a until it becomes practically useless. Linearization dy = y − 2x + x 2 − 4, y (0) = 4 dx a=0 b Euler’s Method Consider the IVP dy dx = f (x, y ), y (a) = y0 , for which you would like to approximate y (b). Let n (the number of steps) be a natural number. 1. Let h = b−a n (the step size). 2. Let x0 = a and xn = b, and recall y0 = y (a) is your given intial value. 3. Let xi = xi−1 + h for 1 ≤ i ≤ n. 4. Let yi = yi−1 + f (xi−1 , yi−1 ) · h. 5. Then, y (b) ≈ yn . Euler’s Method dy = y − 2x + x 2 − 4, y (0) = 4 n = 4, b = 2 dx n=4 a=0 b=2 h = b−a 4 = 0.5 x0 = a = 0 x1 = x0 + h = 0.5 x2 = x1 + h = 1 x3 = x2 + h = 1.5 x4 = x3 + h = 2 = b y0 y1 y2 y3 y4 = y (x0 ) = y (0) = 4 = y0 + f (x0 , y (x0 )) · h = 4 = y1 + f (x1 , y (x1 )) · h = 3.625 = y2 + f (x2 , y (x2 )) · h = 2.9375 = y3 + f (x3 , y (x3 )) · h = 2.03125 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 f (xn , yn ) 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 f (xn , yn ) f (x0 , y0 ) = 0 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 f (xn , yn ) f (x0 , y0 ) = 0 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 f (xn , yn ) f (x0 , y0 ) = 0 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 y2 = 3.625 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 y2 = 3.625 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 f (x2 , y2 ) = −1.375 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 y2 = 3.625 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 f (x2 , y2 ) = −1.375 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 y3 = 2.9375 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 y2 = 3.625 y3 = 2.9375 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 f (x2 , y2 ) = −1.375 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 y3 = 2.9375 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 y2 = 3.625 y3 = 2.9375 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 f (x2 , y2 ) = −1.375 f (x3 , y3 ) = −1.8125 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 y3 = 2.9375 Organizing Information in a Chart dy = y − 2x + x 2 − 4, y (0) = 4 dx n = 4, b = 2 First, we note that a = 0, b = 2, and n = 4. So, h = Also, let f (x, y ) = y − 2x + x 2 − 4. n 0 1 2 3 xn x0 = 0 x1 = 0.5 x2 = 1 x3 = 1.5 yn y0 = 4 y1 = 4 y2 = 3.625 y3 = 2.9375 f (xn , yn ) f (x0 , y0 ) = 0 f (x1 , y1 ) = −0.75 f (x2 , y2 ) = −1.375 f (x3 , y3 ) = −1.8125 2−0 4 = 0.5. yn+1 = yn + f (xn , yn ) · h y1 = 4 y2 = 3.625 y3 = 2.9375 y4 = 2.03125 Example Use Euler’s Method with n = 4 to approximate y (1) given the IVP. Round your answer to three decimal places. dy = arctan(y + x), y (0) = 0 dx (I got y (1) ≈ 0.378)