The exact solution can also be found for the linear equation. Write the answer as a function of . с y(x) = x-1+- Thus the actual value of the function at the point x = 1.1 is y(1.1) =

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Chapter2: Second-order Linear Odes
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The exact solution can also be found for the linear equation. Write the answer as a function of .
C
y(x) = x-1+
Thus the actual value of the function at the point x = 1.1 is
y(1.1) =
Transcribed Image Text:The exact solution can also be found for the linear equation. Write the answer as a function of . C y(x) = x-1+ Thus the actual value of the function at the point x = 1.1 is y(1.1) =
Suppose that we use the Improved Euler's method to approximate the solution to the differential equation
dy
da
y(0.1) = 6.
= x - ly;
Let f(x, y) = x-ly.
We leto = 0.1 and yo = 6 and pick a step size h = 0.25. The improved Euler method is the the following algorithm. From (syn), our
approximation to the solution of the differential equation at the n-th stage, we find the next stage by computing the x-step +1 = ïn +h,
and then k₁, the slope at (n. Yn). The predicted new value of the solution is n+1 = yn + h · k₁. Then we find the slope at the predicted
new point k₂ = f(n+1, %n+1) and get the corrected point by averaging slopes
h
Yn+1 = Yn+(k₁ +k₂).
Complete the following table:
In
In
Yn
k₁
-5.9
k₂
-4.175
0 0.1
6
1 0.35
4.740625
-4.390625
-3.04296875
0.6
3.811425781 -3.211425781 3.008569336
-2.158569336
3 0.85 3.140176391 -2.290176391 2.567632293 -1.467632293
1.1
2.670450306
Zn+1
4.525
3.64296875
Transcribed Image Text:Suppose that we use the Improved Euler's method to approximate the solution to the differential equation dy da y(0.1) = 6. = x - ly; Let f(x, y) = x-ly. We leto = 0.1 and yo = 6 and pick a step size h = 0.25. The improved Euler method is the the following algorithm. From (syn), our approximation to the solution of the differential equation at the n-th stage, we find the next stage by computing the x-step +1 = ïn +h, and then k₁, the slope at (n. Yn). The predicted new value of the solution is n+1 = yn + h · k₁. Then we find the slope at the predicted new point k₂ = f(n+1, %n+1) and get the corrected point by averaging slopes h Yn+1 = Yn+(k₁ +k₂). Complete the following table: In In Yn k₁ -5.9 k₂ -4.175 0 0.1 6 1 0.35 4.740625 -4.390625 -3.04296875 0.6 3.811425781 -3.211425781 3.008569336 -2.158569336 3 0.85 3.140176391 -2.290176391 2.567632293 -1.467632293 1.1 2.670450306 Zn+1 4.525 3.64296875
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