Runge Kutta Method MCQ Quiz in తెలుగు - Objective Question with Answer for Runge Kutta Method - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept: by Runge-Kutta Method

y₁ = y₀ + k

$\mathrm{k}=\frac{1}{6}({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4)$

Calculation:

k₁ = h f(x₀, y₀)

k₁ = 0.1 × [f(x₀, y₀)]

x₀ = 0, y₀ = 1

$\begin{array}{l}\mathrm{f}\left(0,1\right)=\frac{1}{{\mathrm{e}}^{\prime }}=\frac{1}{\mathrm{e}}\\ {\mathrm{k}}_1=\frac{0.1}{\mathrm{e}}=0.037\end{array}$

$\begin{array}{l}{\mathrm{k}}_2=\mathrm{h}\mathrm{f}\left({\mathrm{x}}_0+\frac{\mathrm{h}}{2},{\mathrm{y}}_0+\frac{{\mathrm{k}}_1}{2}\right)\\ \mathrm{f}\left(0+0.05,1+\frac{0.037}{2}\right)\end{array}$

f(0.05, 1.018)

${\mathrm{k}}_2=0.1\times \left[\frac{1}{{\mathrm{e}}^{\left(0.05+1.018\right)}}\right]$

K₂ = 0.034

$\begin{array}{l}{\mathrm{k}}_3=\mathrm{h}\mathrm{f}\left({\mathrm{x}}_0+\frac{\mathrm{h}}{2},{\mathrm{y}}_0+\frac{{\mathrm{k}}_2}{2}\right)\\ \mathrm{f}\left(0.05,1+\frac{0.034}{2}\right)=\mathrm{f}\left(0.05,1.017\right)\\ {\mathrm{k}}_3=0.1\times \left[\frac{1}{{\mathrm{e}}^{\left(0.05+1.017\right)}}\right]\end{array}$

K₂ = 0.034

${\mathrm{k}}_4=\mathrm{h}\mathrm{f}({\mathrm{x}}_0+\mathrm{h},{\mathrm{y}}_0+{\mathrm{k}}_3)$

$\begin{array}{l}{\mathrm{k}}_4=\mathrm{h}\mathrm{f}\left(0.05,1.034\right)\\ {\mathrm{k}}_4=0.1\times \left[\frac{1}{{\mathrm{e}}^{\left(0.05+1.034\right)}}\right]\end{array}$

K₄ = 0.034

$\mathrm{k}=\frac{1}{6}\left[0.037+2\times 0.034+2\times 0.034+0.034\right]$

k = 0.0345

y₁ = 1 + 0.0345

From Runge Kutta 3^rd order,

${\mathrm{y}}_1={\mathrm{y}}_0+\frac{1}{6}\left({\mathrm{k}}_1+4{\mathrm{k}}_2+{\mathrm{k}}_3\right)$

${\mathrm{k}}_1=\mathrm{h}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=0.1\left[2{\mathrm{x}}_0+{\mathrm{y}}_0\right]=0.1\left[0+1\right]=0.1$

${\mathrm{k}}_2=\mathrm{h}\mathrm{f}\left({\mathrm{x}}_0+\frac{\mathrm{h}}{2},{\mathrm{y}}_0+\frac{{\mathrm{k}}_1}{2}\right)=0.1\mathrm{f}\left(0.05,1+\frac{0.1}{2}\right)=0.1\mathrm{f}\left(0.05,1.05\right)$

$=0.1\left[2\left(0.05\right)+1.05\right]=0.115$

${\mathrm{k}}_3=\mathrm{h}\mathrm{f}\left({\mathrm{x}}_0+\mathrm{h},{\mathrm{y}}_0+2{\mathrm{k}}_2-{\mathrm{k}}_1\right)$

$=0.1\mathrm{f}\left(0.1,1+0.23-0.1\right)$

$=0.1\mathrm{f}(0.1,1.13)$

$=0.1[2(0.1)+1.13]=0.133$

$\frac{dy}{dx}=\left(x+y\right)e^x=f\left(x,y\right)$

given that

$x_0=0,y_0=1,h=0.1$

i) $k_1=hf\left(x_0,y_0\right)=h\left[\left(x_0+y_0\right)e^{x_0}\right]$

$=0.1\left[\left(0+1\right)e^0\right]=0.1$

ii) $k_2=hf\left(x_0+h,y_0+k_1\right)=h\left[\left(x_0+h,y_0+k_1\right)e^{x_0+h}\right]$

$=0.1\left[\left(0+0.1+1+0.1\right)e^{\left(0+0.1\right)}\right]=0.13262$

then $k=\frac{k_1+k_2}{2}=\frac{0.13262+0.1}{2}=0.11631$

so $y_1=y_o+k\Rightarrow y\left(0.1\right)=1+.11631$

$\mathrm{y}(0.1)=1.11631$

II^nd approximation

${\mathrm{x}}_1={\mathrm{x}}_0+\mathrm{h}=0.1,{\mathrm{y}}_1=1.11631,\mathrm{h}=0.1$

i) $k_1=hf\left(x_1+y_1\right)=0.1\left[\left(0.1+1.11631\right)e^{0.1}\right]=0.134423$

ii) $k_2=hf\left(x_1+h_1+y_1+k_1\right)=0.1\left[\left(0.1+0.1+1.11631+0.1344230\right)e^{\left(0.1+0.1\right)}\right]$

$=0.177192$

$k=\frac{k_1+k_2}{2}=\frac{.134423+0.177192}{2}=0.155807$

$\begin{array}{l}{y}_2=y_1+k\Rightarrow y\left(0.2\right)=1.11631+0.155807\\ y\left(0.2\right)=1.272117\end{array}$

$\frac{dy}{dx}=\left(x+y\right)e^x$

given that

$x_0=0,y_0=1,h=0.1$

i) $k_1=hf\left(x_0,y_0\right)=h\left[\left(x_0+y_0\right)e^{x_0}\right]$

$=0.1\left[\left(0+1\right)e^0\right]=0.1$

ii) $k_2=hf\left(x_0+h,y_0+k_1\right)=h\left[\left(x_0+h,y_0+k_1\right)e^{x_0+h}\right]$

$=0.1\left[\left(0+0.1+1+0.1\right)e^{(0+0.1)}\right]=0.13262$

then $k=\frac{k_1+k_2}{2}=\frac{0.13262+0.1}{2}=0.11631$

so $y_1=y_o+k\Rightarrow y\left(0.1\right)=0+.11631=0.11631$

II^nd approximation

x₁ = 0.1, h = 0.1, y₁ = 0.11631

i) $k_1=hf\left(x_1+y_1\right)=0.1\left[\left(0.1+0.11631\right)e^{0.1}\right]=0.02390$

ii) $k_2=hf\left(x_1+h,y_1+k_1\right)=0.1\left[\left(0.1+0.1+.11631+.02390\right)e^{0.1+0.1}\right]$

= 0.04155

$\begin{array}{l}k=\frac{k_1+k_2}{2}=\frac{.02390+0.04155}{2}=.032725\end{array}$

$$\begin{gathered} y_2=y_1+k\Rightarrow y\left(0.2\right)=.11631+.032725 \\ y\left(0.2\right)=.149035 \end{gathered}$$

$\begin{array}{l}{y}_1=y_0+\frac{1}{2}\left(k_1+k_2\right)\\ k_1=hf\left(x_0,y_0\right)=0.1\left(0-0\right)=0\\ k_2=hf\left(x_0+h,y_0+k_1\right)=0.1\left(0.1-0\right)=0.01\\ \therefore y_1=0+\frac{1}{2}\left(0+0.01\right)=0.005\end{array}$