Question

Consider an ordinary differential equation $\frac{dx}{dt}=4t+4$. If $x=x_{0}att=0$, the increment in x calculated using Runge-Kutta fourth order multi-step method with a step size of $\Delta t=0.2$ is

• A. 0.22
• B. 0.44
• C. 0.66
• D. 0.88

Answer 1

The correct option is D 0.88

$\frac{dx}{dt}=4t+4=f(t,x)$
$h=\Delta t=0.2,t_0=0$
$k_1=\Delta tf(0,x_0)=0.2(0+4)=0.8$
$k_2=f(t_0+\frac{\Delta t}{2},x_0+\frac{1}{2}{k}_1)\Delta t$
$=\left[f\right(0.1,x_0+0.4\left)\right]\Delta t$
$=\left[4\right(0.1)+4]\left(0.2\right)$
$=0.88$
$k_3=f(t_0+\frac{\Delta t}{2},x_0+\frac{1}{2}{k}_2)\Delta t$
$=\left[f\right(0.1,x_0+0.44\left)\right]\left(0.2\right)$
$=\left[4\right(0.1)+4]\left(0.2\right)$
$=0.88$
$k_4=f(t_0+\Delta t,x_0+k_3)\Delta t$
$=\left[f\right(0.2,x_o+0.88\left)\right]\Delta t$
$=[4\times 0.2+4]\left(0.2\right)$
$=0.96$
$x_1=x\left(0.2\right)$
$=x_0+\frac{1}{6}[k_1+2k_2+2k_3+k_4]$
$=x_0+\frac{1}{6}[0.8+2\times 0.88+2\times 0.88+0.96$
$=x_0+\frac{0.2}{6}[4+2\times 4.4+2\times 4.4+4.8]$
$x_1-x_0=0.88$
$\mathrm{Increment}\mathrm{in}xis0.88$