Answer to Question #257973 in Calculus for sagun poudel

Question

Solve the following Initial Value Problem using (i)R-K method of O(h²)

and (ii) R-K method of O(h⁴)

y' = x²y + x³ and y(0) = 1.

Find y(0.4) taking h = 0.2, where y' means dy/dx

Answer 1

(i) Rung-Kutta method of second orde

$y'=x^2y+x^3=\frac{dy}{dx}$

$h=0.2$

$y(0)=1, y_0=1, x_0=0$

$f(x,y)=x^2y+x^3, u_1=u_0+h=0+0.2\implies u_1=0.2$

Applying Runge-Kutta order 2,

$k_1=hf(x_0,y_0)$

$=0.2f(0,1)=0.2(0^2+0^3)=0$

$k_2=hf(x_0+h,y_0+k_1)=hf(0.2,1)$

$=0.2(0.2^2\cdot1+0.2^3)=0.0096$

$\therefore y_1=y(0.2)=y_0+\frac{1}{2}(k_1+k_2)$

$=1+\frac{1}{2}(0+0.0096)=1.0048$

Now; $x_1=0.2, y_1=1.0048,h=0.2$

$\therefore u_2=u_1+h=0.2+0.2=0.4$

$f(x,y)=x^2y+x^3$

Applying R-K method order 2

$k_1=hf(x_1,y_1)=0.2f(0.2,1.0048)$

$=0.2((0.2)^2\cdot1.0048+(0.2)^3)=0.0096$

$k_2=hf(x_1+h,y_1+k)$

$=hf(0.4,1.0144)$

$=(0.2)((0.4)^2\cdot1.0144+(1.0144)^3)=0.2412$

$y_2=y(0.4)=y_1+\frac{1}{2}(k_1+k_2)$

$=1.0048+\frac{1}{2}(0.0096+0.2412)$

$=1.1302$

(ii) $f(x,y)=x^2y+x^3$

$h=0.2$

$y(0)=1, y_0=1, x_0=0$

Applying Runge-Kutta method of fourth order,

$k_1=hf(x_0,y_0)=0.2f(0,1)$

$=0.2(0+0)=0$

$k_2=hf(x_0+\frac{h}{2},y_0+\frac{k}{2})=hf(0.1,1)$

$=0.2(0.1^2\cdot1+(0.1)^3)=0.0022$

$k_3=hf(x_0+\frac{h}{2},y_0+\frac{k_1}{2})=hf(0.1,1.0011)$

$=0.2((0.1)^2\cdot(1.0011)+(0.1)^3)=0.0022$

$k_4=hf(x_0+h,y_0+k_3)=hf(0.2,1.0022)$

$=(0.2)((0.2)^2\cdot(1.0022)+(0.2)^3)=0.0096$

$k=\frac{1}{6}(k_1+2k_2+2k_3+k_4)=0.0031$

$\therefore y_1=y(0.2)=y_0+k=1+0.0031=1.0031$

Now $x_1=0.2,y_1=1.0031,h=0.2$

$f(x,y)=x^2y+x^3$

$x_2=x_1+h=0.2+0.2=0.4$

By applying R-K method of order 4,

$k_1=hf(x_1,y_1)=0.2f(0.2,1.0031)$

$=(0.2)((0.2)^2\cdot(1.0031)+(0.2)^3)$

=0.0096

$k_2=hf(x_1+\frac{h}{2},y_1+\frac{k_1}{2})=hf(0.3,1.0079)$

$=(0.2)((0.3)^2\cdot(1.0079)+(0.3)^3)=0.0235$

$k_3=hf(x_1+\frac{h}{2},y_1+\frac{k_2}{2})=hf(0.3,1.0148)$

$=(0.2)((0.3)^2\cdot(1.0148)+(0.3)^3)=0.0237$

$k_4=hf(x_1+h,y_1+k_3)=hf(0.4,1.0268)$

$=(0.2)((0.4)^2\cdot(1.0268)+(0.4)^3)=0.0456$

$k=\frac{1}{6}(k_1+2k_2+2k_3+k_4)$

$=0.0249$

$\therefore y_2=y(0.4)=y_1+k$

$=1.0031+0.0249$

$=1.028$