Question

Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4. (A computer algebra system is recommended. Round your answers to five decimal places.) y' = 9 + t - y, y(0) = 7 (a) Use the improved Euler method with h = 0.05. y(0.1) = y(0.2) = y(0.3) = y(0.4) = (b) Use the improved Euler method with h = 0.025. y(0.1) = y(0.2) = y(0.3) = y(0.4) = (c) Use the improved Euler method with h = 0.0125. y(0.1) = y(0.2) = y(0.3) = y(0.4) =

          Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4. (A computer algebra system is recommended. Round your answers to five decimal places.)
y' = 9 + t - y, y(0) = 7
(a) Use the improved Euler method with h = 0.05.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(b) Use the improved Euler method with h = 0.025.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(c) Use the improved Euler method with h = 0.0125.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
        
Show more…
Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4. (A computer algebra system is recommended. Round your answers to five decimal places.)
y' = 9 + t - y, y(0) = 7
(a) Use the improved Euler method with h = 0.05.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(b) Use the improved Euler method with h = 0.025.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(c) Use the improved Euler method with h = 0.0125.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =

Added by Jason L.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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00:01 We have f of t y is equal to y dash equal to 9 plus t minus y and we have y 0 is equal to 7 now euler method is y plus y n plus 1 is equal to y n plus h into f of t n y n here we are using h is equal to 0 .05 so putting n is equal to 0 y 1 is equal to y 0 plus h f t 0 y 0 so this is equal to 7 plus 0 .05 into 9 plus 0 minus 7 which is equal to 7 .1 so we got y of 0 .1 is equal to 7 .1 from here t1 is equal to 0 .1 and y1 is equal to 7 .1 similarly putting n is equal to 1 in euler method we got y2 is equal to y1 plus h times f of t 1 y1 putting all the values we get 7 .1 plus 0 .05 into 9 plus 0 .1…
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