3. 3E: Runge-Kutta Usuli (Mashqlar) Matematika

Exercise 6
Let (y(t)) be a solution to (y' + frac<1><2>,ty^2 = 2) with (y(2) = 1) . Use implicit Euler with a step size (h=1) to estimate an approximate value of (y(5)) . [Let’s focus on the solution that will come up if we take positive square roots at every juncture possible.]
Qaror
The step size is (h=1) and the function (f(t,y)) is given by (2-frac<1><2>,ty^2) . The equation that implicit Euler gives us is [y_ - left( 2 - frac<1> <2>, t_ y_^2 ight) = y_n ,.] We complete the square next. Multiply the equation through by (frac<2><>>) to make the coefficient on (y_^2) equal to one, then add (frac<1><>^2>) to both sides. This gives us [left( y_ + frac<1><>> ight)^2 = frac<2y_n><>> + frac<4><>> + frac<1><>^2> ,,] so we’ve written (y_) in terms of stuff we can handle when we say [y_ = sqrt< frac<2y_n><>> + frac<4><>> + frac<1><>^2> > - frac<1><>> ,.] Now we fill in the table: [egin t & 2 & 3 & 4 & 5 hline y & 1 & 1.1196 & 1.0237 & 0.9178 end] Our estimate is (y(5) approx 0.9178) .
Exercise 7
Let (y(t)) be the solution to the initial-value problem (y' = sqrt) , (y(0) = 3) . Using four steps, estimate (y(1)) with the implicit Euler method. [Take positive square roots should the need arise, and use four decimals of precision.] [Hint. Though it doesn’t look like it at first, (y + alphasqrt) is the sort of thing that’s amenable to completing the square, where (alpha) is some constant.]
Qaror
We’ll need to do some algebra before we can plug in any numbers. The equation governing the next estimate of the function is [y_ - frac<1> <4>, sqrt<>> = y_n ,.] By adding (frac<1><64>) to both sides, we can complete a non-obvious square. This gives us [left( sqrt<>> - frac<1> <8> ight)^2 = y_n + frac<1> <64>,,] so the relation we went looking for is in fact [y_ = left( sqrt<64>> + frac<1> <8> ight)^2 ,.] Armed with this, we can make short work of the running of the algorithm: [egin t & 0 & 0.25 & 0.5 & 0.75 & 1 hline y & 3 & 3.4654 & 3.9631 & 4.4930 & 5.0551 end]
So our estimate is (y(1) approx 5.0551) .