12.
Explanation
The initial value problem using the 4th order Runge-Kutta method with a step size of 0.5, we'll perform iterations to approximate the values of y at discrete points within the interval [0, 1].
The differential equation is given as:
y′=y3x−xy
with the initial condition 2y(0)=2.
We'll use the following formula to update the value of y at each step:
k1=h⋅f(xi,yi)
k2=h⋅f(xi+2h,yi+2k1)
k3=h⋅f(xi+2h,yi+2k2)
k4=h⋅f(xi+h,yi+k3)
yi+1=yi+61(k1+2k2+2k3+k4)
Let's calculate the values of y at each step within the interval [0, 1]:
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Initial condition: y0=2
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ℎ=0.5h=0.5
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Number of steps (n) = 1−00.50.51−0 = 2
Step 1: 0=0x0=0
(02−0⋅2)=0k1=0.5⋅(y03x0−x0⋅y0)=0.5⋅(20−0⋅2)=0 (0.752.125−0.25⋅2.25)=−0.03529k2=0.5⋅((y0+0.25⋅0.5)3(x0+0.25)−(x0+0.25)⋅(y0+0.25))=0.5⋅(2.1250.75−0.25⋅2.25)=−0.03529
((y0+0.25⋅(−0.03529))3(x0+0.25)−(x0+0.25)⋅(y0+0.25⋅(−0.03529)))=−0.1512
k4=0.5⋅((y0+0.5⋅(−0.1512))3(x0+0.5)−(x0+0.5)⋅(y0+0.5⋅(−0.1512)))=−0.3795
31y1=2+61(0+2⋅(−0.03529)+2⋅(−0.1512)+(−0.3795))=1.631
Step 2: 5x1=0.5
k1=0.5⋅(y13x1−x1⋅y1)=0.5⋅(1.6310.75−0.5⋅1.631)=0.2319
k2=0.5⋅((y1+0.25⋅0.2319)3(x1+0.25)−(x1+0.25)⋅(y1+0.25⋅0.2319))=0.1062
k3=0.5⋅((y1+0.25⋅0.1062)3(x1+0.25)−(x1+0.25)⋅(y1+0.25⋅0.1062))=0.04886
k4=0.5⋅((y1+0.5⋅0.04886)3(x1+0.5)−(x1+0.5)⋅(y1+0.5⋅0.04886))=0.1847
y2=1.631+61(0.2319+2⋅0.1062+2⋅0.04886+0.1847)=1.694
So, using the 4th order Runge-Kutta method with a step size of 0.5, the approximate solution for the initial value problem y′=y3x−xy, (0)=2 over the interval [0, 1] is 1≈1.694y(1)≈1.694.