Question

Answer 1

The second order differential equation is given as:

$\frac{\mathrm{d} ^2y}{\mathrm{d} t^2}+2\left ( \frac{\mathrm{d} y}{\mathrm{d}t } \right )^2-7y=1.2, \ \ y(0)=4,y'(0)=0$

$t=0.2, \ h=0.1$

Now we have to convert the given second order differential equation to a set of two first order differential equations.

Let $u(t)=y(t)$ and $v(t)=y'(t)$

Then,

$u'=y'=v$

$v'=y''=1.2+7y-2y'^2=1.2+7u-2v^2$

Initial conditions are given by:

$u(0)=y(0)=4$

$v(0)=y'(0)=0$

i.e the two 1st order differential equations are:

$u'=v$

$v'=1.2+7u-2v^2$

with initial conditions $u(0)=4 \ and \ v(0)=0$

Now we use Runge-Kutta method to solve the problem,

$u'=f(u,v)=v,\ u(0)=4$

$v'=g(u,v)=1.2+7u-2v^2 , \ v(0)=0$

From t= 0 to 0.2 it takes 2 steps. i.e $t_0=0, t_1=0.1 , t_2=0.2$

Step 1:   $n=0, u_0=4,v_0=0$ we have,

$k_1=hf(u_n,v_n)=0.1f(4,0)=0$

$l_1=hg(u_n,v_n)=0.1g(4,0)=0.1(29.2)=2.92$

$k_2=hf(u_0+\frac{1}{2}k_1,v_0+\frac{1}{2}l_1)=0.1f(4,1.46)=0.146$

$l_2=hg(u_0+\frac{1}{2}k_1,v_0+\frac{1}{2}l_1)=0.1g(4,1.46)=2.494$

$k_3=hf(u_0+2k_2-k_1,v_0+2l_2-l_1)=0.1f(4.29,2.07)=0.207$

$l_3=hg(u_0+2k_2-k_1,v_0+2l_2-l_1)=0.1g(4.29,2.07)=2.26$

$u_{n+1}=u_1=u_0+\frac{1}{6}(k_1+4k_2+k_3)=4+\frac{1}{6}(0.791)=4.132=u(t_1)$

$t_1=t_0+h=0+0.1=0.1$

$\therefore u(0.1)=4.132$

$v_{n+1}=v_1=v_0+\frac{1}{6}(l_1+4l_2+l_3)=\frac{1}{6}(7.674)=1.279$

$\therefore v(0.1)=1.279$

Step 2: $n=1, u_1=4.132,v_1=1.279$

$k_1=hf(u_1,v_1)=0.1f(4.132,1.279)=0.1279$

$l_1=hg(u_1,v_1)=0.1g(4.132,1.279)=2.685$

$k_2=hf(u_1+\frac{1}{2}k_1,v_1+\frac{1}{2}l_1)=0.1f(4.195,2.622)=0.2622$

$l_2=hg(u_1+\frac{1}{2}k_1,v_1+\frac{1}{2}l_1)=0.1g(4.195,2.622)=1.682$

$k_3=hf(u_1+2k_2-k_1,v_1+2l_2-l_1)=0.1f(4.528,1.958)=0.1958$

$l_3=hg(u_1+2k_2-k_1,v_1+2l_2-l_1)=0.1g(4.528,1.958)=2.522$

$u_{n+1}=u_2=u_1+\frac{1}{6}(k_1+4k_2+k_3)=4.132+\frac{1}{6}(1.373)=4.361=u(t_2)$

$t_2=t_1+h=0.1+0.1=0.2$

$\therefore u(0.2)=4.361$

$v_{n+1}=v_2=v_1+\frac{1}{6}(l_1+4l_2+l_3)=1.279+1.989=3.268$

$\therefore v(0.2)=3.268$