Answer to Question #283846 in Quantitative Methods for Ar-Ar

Question

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.

Answer 1

Solution:

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

$\dfrac{dy}{dx}=yx^2-1.2y$

(1) Analytically:

$\\ \Rightarrow \dfrac{dy}{dx}=y(x^2-1.2) \\ \Rightarrow \dfrac{dy}{y}=(x^2-1.2)dx \\\Rightarrow \log y=\dfrac{x^3}3-1.2x+C$

y(0)=1

When x=0,y=1

Then,

$\log 1=0+C \\\Rightarrow C=0$

So, $\log y=\dfrac{x^3}3-1.2x$

Or, $y=e^{\frac{x^3}3-1.2x}$

(2) Using Euler’s method.

$\begin{aligned}&\text { (a) } \mathrm{h}=0.5 . \quad \mathrm{y}_{\mathrm{i}+1}=\mathrm{y}_{\mathrm{i}}+\mathrm{f}\left(\mathrm{x}_{\mathrm{i}}, \mathrm{y}_{\mathrm{i}}\right) \mathrm{h} \quad \text { where } \mathrm{f}(\mathrm{x})=\mathrm{yx}^{2}-1.2 \mathrm{y} \\&\mathrm{i}=0 \quad \mathrm{y}_{1}=\mathrm{y}_{0}+\mathrm{f}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right) \mathrm{h} \rightarrow \quad \mathrm{y}_{1}=\mathrm{y}(0.5)=1+\left(1 * 0^{2}-1.2 * 1\right) 0.5=0.4 \\&\mathrm{i}=1 & \mathrm{y}_{2}=\mathrm{y}_{1}+\mathrm{f}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \mathrm{h} \rightarrow \mathrm{y}_{2}=\mathrm{y}(1.0)=0.4+\left(0.4 * 0.5^{2}-1.2 * 0.4\right) 0.5=0.21 \\&\mathrm{i}=2 & \mathrm{y}_{3}=\mathrm{y}_{2}+\mathrm{f}(\mathrm{x}_{2}, \mathrm{y}_{2}) \mathrm{h} \rightarrow \mathrm{y}_{3}=\mathrm{y}(1.5)=0.21+(0.21 * 1^{2}-1.2 * 0.21) 0.5=0.189 \\&\mathrm{i}=3 & \mathrm{y}_{4}=\mathrm{y}_{3}+\mathrm{f}(\mathrm{x}_{3}, \mathrm{y}_{3} \mathrm{h} \rightarrow \mathrm{y}_{4}=\mathrm{y}(2.0)=0.189+\left(0.189 * 1.5^{2}-1.2 * 0.189\right) 0.5=0.2882\end{aligned} -$

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.

$\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}$