Question #283846

Solve the following ordinary differential equation over the interval from x = 0 to 1 using a



step size of 0.25 where y(0) = 1.



𝒅𝒚



𝒅𝒙 = (𝟏 + 𝟐𝒙)𝒚



𝟐



(1) Analytically.



(2) Using Euler’s method.



(3) Using Heun’s method without iteration.



(4) Using the fourth-order RK method.

1
Expert's answer
2022-01-02T16:48:01-0500

Solution:

Since, the given differential equation is unclear, we assume it as follows:

dydx=yx21.2y\dfrac{dy}{dx}=yx^2-1.2y

(1) Analytically:

dydx=y(x21.2)dyy=(x21.2)dxlogy=x331.2x+C\\ \Rightarrow \dfrac{dy}{dx}=y(x^2-1.2) \\ \Rightarrow \dfrac{dy}{y}=(x^2-1.2)dx \\\Rightarrow \log y=\dfrac{x^3}3-1.2x+C

y(0)=1

When x=0,y=1

Then,

log1=0+CC=0\log 1=0+C \\\Rightarrow C=0

So, logy=x331.2x\log y=\dfrac{x^3}3-1.2x

Or, y=ex331.2xy=e^{\frac{x^3}3-1.2x}

(2) Using Euler’s method.

 (a) h=0.5.yi+1=yi+f(xi,yi)h where f(x)=yx21.2yi=0y1=y0+f(x0,y0)hy1=y(0.5)=1+(1021.21)0.5=0.4i=1y2=y1+f(x1,y1)hy2=y(1.0)=0.4+(0.40.521.20.4)0.5=0.21i=2y3=y2+f(x2,y2)hy3=y(1.5)=0.21+(0.21121.20.21)0.5=0.189i=3y4=y3+f(x3,y3hy4=y(2.0)=0.189+(0.1891.521.20.189)0.5=0.2882\begin{aligned}&\text { (a) } \mathrm{h}=0.5 . \quad \mathrm{y}_{\mathrm{i}+1}=\mathrm{y}_{\mathrm{i}}+\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}, \mathrm{y}_{\mathrm{i}}\right) \mathrm{h} \quad \text { where } \mathrm{f}(\mathrm{x})=\mathrm{yx}^{2}-1.2 \mathrm{y} \\&\mathrm{i}=0 \quad \mathrm{y}_{1}=\mathrm{y}_{0}+\mathrm{f}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right) \mathrm{h} \rightarrow \quad \mathrm{y}_{1}=\mathrm{y}(0.5)=1+\left(1 * 0^{2}-1.2 * 1\right) 0.5=0.4 \\&\mathrm{i}=1 & \mathrm{y}_{2}=\mathrm{y}_{1}+\mathrm{f}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \mathrm{h} \rightarrow \mathrm{y}_{2}=\mathrm{y}(1.0)=0.4+\left(0.4 * 0.5^{2}-1.2 * 0.4\right) 0.5=0.21 \\&\mathrm{i}=2 & \mathrm{y}_{3}=\mathrm{y}_{2}+\mathrm{f}(\mathrm{x}_{2}, \mathrm{y}_{2}) \mathrm{h} \rightarrow \mathrm{y}_{3}=\mathrm{y}(1.5)=0.21+(0.21 * 1^{2}-1.2 * 0.21) 0.5=0.189 \\&\mathrm{i}=3 & \mathrm{y}_{4}=\mathrm{y}_{3}+\mathrm{f}(\mathrm{x}_{3}, \mathrm{y}_{3} \mathrm{h} \rightarrow \mathrm{y}_{4}=\mathrm{y}(2.0)=0.189+\left(0.189 * 1.5^{2}-1.2 * 0.189\right) 0.5=0.2882\end{aligned} -

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%\begin{aligned} &\mathrm{y}_{\mathrm{i}+1}=\mathrm{y}_{\mathrm{i}}+\mathrm{h}\left(\mathrm{k}_{\mathbf{1}}+2 \mathrm{k}_{\mathbf{2}}+2 \mathrm{k}_{\mathbf{3}}+\mathrm{k}_{\mathbf{4}}\right) / 6 \\ &\mathrm{k}_{\mathbf{1}}=\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}, \mathrm{y}_{\mathrm{i}}\right) \quad \mathrm{k}_{\mathbf{2}}=\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}+\mathrm{h} / 2, \mathrm{y}_{\mathrm{i}}+\mathrm{k}_{\mathbf{1}} \mathrm{h} / 2\right) \quad \mathrm{k}_{\mathbf{3}}=\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}+\mathrm{h} / 2, \mathrm{y}_{\mathrm{i}}+\mathrm{k}_{\mathbf{2}} \mathrm{h} / 2\right) \quad \mathrm{k}_{\mathbf{4}}=\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}+\mathrm{h}, \mathrm{y}_{\mathrm{i}}+\mathrm{k}_{\mathbf{3}} \mathrm{h}\right) \\ &\mathrm{i}=0, \mathrm{k}_{\mathbf{1}}=\mathrm{f}(0,1)=-1.2, \mathrm{k}_{\mathbf{2}}=\mathrm{f}(0.25,0.7)=-0.79625 \\ &\mathrm{k}_{\mathbf{3}}=\mathrm{f}(0.25,0.800938)=-0.91107, \quad \mathrm{k}_{\mathbf{4}}=\mathrm{f}(0.5,0.544467)=-0.51724 \\ &\mathrm{y}_{\mathbf{1}}=\mathrm{y}(0.5)=\mathbf{0 . 5 7 2 3 4 4} \quad \varepsilon_{\mathrm{t}}=\mathbf{0 . 0 3 2} \% \\ &\mathrm{i}=1, \quad \mathrm{k}_{\mathbf{1}}=\mathrm{f}(0.5,0.572344)=-0.54373, \mathrm{k}_{\mathbf{2}}=\mathrm{f}(0.75,0.436412)=-0.27821 \\ &\mathrm{k}_{\mathbf{3}}=\mathrm{f}(0.75,0.50279)=-0.32053, \mathrm{k}_{\mathbf{4}}=\mathrm{f}(1,0.412079)=-0.08242 \\ &\mathrm{y}_{\mathbf{2}}=\mathrm{y}(1)=\mathbf{0 . 4 2 0 3 7 5} \quad \varepsilon_{\mathrm{t}}=\mathbf{0 . 0 0 6} \% \\ &\mathrm{i}=2, \quad \mathrm{k}_{\mathbf{1}}=\mathrm{f}(1,0.420375)=-0.08407, \mathrm{k}_{\mathbf{2}}=\mathrm{f}(1.25,0.399356)=0.144767 \\ &\mathrm{k}_{\mathbf{3}}=\mathrm{f}(1.25,0.456567)=0.165505, \mathrm{k}_{\mathbf{4}}=\mathrm{f}(1.5,0.503128)=0.528284 \\ &\mathrm{i}_{\mathbf{3}}=\mathrm{y}(1.5)=\mathbf{0 . 5 0 9 1 0 4} \quad \varepsilon_{\mathrm{t}}=\mathbf{0 . 0 1} \% \\ &\mathrm{k}_{\mathbf{1}}=\mathrm{f}(1.5,0.509104)=0.534559, \mathrm{k}_{\mathbf{2}}=\mathrm{f}(1.75,0.642744)=1.197111 \\ &\mathrm{k}_{\mathbf{3}}=\mathrm{f}(1.75,0.808382)=0.505611, \mathrm{k}_{\mathbf{4}}=\mathrm{f}(2,1.26191)=3.533348 \\ &\mathrm{y}_{\mathbf{4}}=\mathrm{y}(2)=\mathbf{1} .29855 \quad \varepsilon_{\mathrm{t}}=\mathbf{0 . 5 4} \% \end{aligned}


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