Since, the given differential equation is unclear, we assume it as follows:
dxdy=yx2−1.2y
(1) Analytically:
⇒dxdy=y(x2−1.2)⇒ydy=(x2−1.2)dx⇒logy=3x3−1.2x+C
y(0)=1
When x=0,y=1
Then,
log1=0+C⇒C=0
So, logy=3x3−1.2x
Or, y=e3x3−1.2x
(2) Using Euler’s method.
(a) h=0.5.yi+1=yi+f(xi,yi)h where f(x)=yx2−1.2yi=0y1=y0+f(x0,y0)h→y1=y(0.5)=1+(1∗02−1.2∗1)0.5=0.4i=1i=2i=3y2=y1+f(x1,y1)h→y2=y(1.0)=0.4+(0.4∗0.52−1.2∗0.4)0.5=0.21y3=y2+f(x2,y2)h→y3=y(1.5)=0.21+(0.21∗12−1.2∗0.21)0.5=0.189y4=y3+f(x3,y3h→y4=y(2.0)=0.189+(0.189∗1.52−1.2∗0.189)0.5=0.2882−
Analytical solution is shown with blue.
Solution for h=0.5 is shown with red.
(3) Using Heun’s method without iteration.
h = 0.5 , f(x) = yx2 – 1.2y , y0 i+1 = yi + f(xi , yi ) h , yk i+1 = yi + [f(xi , yi ) + f(xi+1, yk-1 i+1)]/2 * h
i = 0 Predictor: y0 1 = 1 + (1*02 – 1.2*1) 0.5 = 0.4
Corrector:
k=1 y 1 1 = 1 + [(1*02 – 1.2*1) + (0.4*0.52 – 1.2*0.4) ]/2 * 0.5 = 0.605 ea= 33.9 %
k=2 y 2 1 = 1 + [(1*02 – 1.2*1) + (0.605*0.52 – 1.2*0.605) ]/2 * 0.5 = 0.5563125 ea= 8.75 %
k=3 y 3 1 = 1 + [(1*02 – 1.2*1) + (0.5563*0.52 – 1.2*0.5563) ]/2 * 0.5 = 0.5678757 ea= 2.04 %
k=4 y 4 1 = 1 + [(1*02 – 1.2*1) + (0.5679*0.52 – 1.2*0.5679) ]/2 * 0.5 = 0.5651295 ea= 0.49 %
y1 = y(0.5) = 0.5651295
i = 1 Predictor: y0 2 = 0.5651295 + (0.5651295*0.52 – 1.2*0.5651295) 0.5 = 0.2966930
Corrector:
k=1 y 1 2 = 0.5651295 + [(05651295*0.52 – 1.2*05651295) + (0.296693*12 – 1.2*0.296693)]/2*0.5
= 0.4160766 ea= 28.7 %
. . . . .
Continue like this to find y(1)=0.4104059, y(1.5)=0.5279021, y(2)=2.181574
(4) Using the fourth-order RK method.
yi+1=yi+h(k1+2k2+2k3+k4)/6k1=f(xi,yi)k2=f(xi+h/2,yi+k1h/2)k3=f(xi+h/2,yi+k2h/2)k4=f(xi+h,yi+k3h)i=0,k1=f(0,1)=−1.2,k2=f(0.25,0.7)=−0.79625k3=f(0.25,0.800938)=−0.91107,k4=f(0.5,0.544467)=−0.51724y1=y(0.5)=0.572344εt=0.032%i=1,k1=f(0.5,0.572344)=−0.54373,k2=f(0.75,0.436412)=−0.27821k3=f(0.75,0.50279)=−0.32053,k4=f(1,0.412079)=−0.08242y2=y(1)=0.420375εt=0.006%i=2,k1=f(1,0.420375)=−0.08407,k2=f(1.25,0.399356)=0.144767k3=f(1.25,0.456567)=0.165505,k4=f(1.5,0.503128)=0.528284i3=y(1.5)=0.509104εt=0.01%k1=f(1.5,0.509104)=0.534559,k2=f(1.75,0.642744)=1.197111k3=f(1.75,0.808382)=0.505611,k4=f(2,1.26191)=3.533348y4=y(2)=1.29855εt=0.54%
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