Chapter 25, Problem 1P

Solve the following initial value problem over the interval from $t=0\text{ to 2 where }y\left(0\right)=1$. Display all your results on the same graph.

$\frac{dy}{dt}=y{t}^2-1.1y$

(a) Analytically.

(b) Euler's method with $h=0.5\text{ and 0}\text{.25}$.

(c) Midpoint method with $h=0.5$.

(d) Fourth-order RK method with $h=0.5$.

To determine

To calculate: The solution of the initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$ with the initial condition as $y\left(0\right)=1$, analytically.

Answer to Problem 1P

Solution:

The solution to the initial value problem is $y\left(t\right)=e^{\frac{t^3}{3}-1.1t}+C$.

Explanation of Solution

Given Information:

The initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$ where $y\left(0\right)=1$.

Formula used:

Tosolve an initial value problem of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$ integrate both sides of the following equation:

$\frac{dy}{h\left(y\right)}=g\left(x\right)dx$

Calculation:

Rewrite the provided differential equation as,

$\begin{array}{c}\frac{dy}{dt}=y{t}^2-1.1y\\ \frac{dy}{dt}=y\left(t^2-1.1\right)\\ \frac{dy}{y}=\left(t^2-1.1\right)dt\end{array}$

Integrate both sides to get,

$\begin{array}{c}\ln y=\frac{t^3}{3}-1.1t+C\\ y\left(t\right)=e^{\frac{t^3}{3}-1.1t}+C\end{array}$

Now use the initial condition $y\left(0\right)=1$ to get the value of constant C.

$\begin{array}{c}y\left(0\right)=e^{\frac{0^3}{3}-1.1\left(0\right)}+C\\ 1=1+C\\ C=0\end{array}$

Hence, the analytical solution of the initial value problem is $y\left(t\right)=e^{\frac{t^3}{3}-1.1t}$.

To determine

To calculate: The solution of the initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$ with the initial condition as $y\left(0\right)=1$ in the interval $t\in \left(0,2\right)$ with step-sizes 0.5 and 0.25 using the Euler’s method.

Answer to Problem 1P

Solution:

For $h=0.5$ the values are,

t	y	$\frac{dy}{dt}$
0	1	$-1.1$
0.5	0.45	$-0.3825$
1	0.25875	$-0.02588$
1.5	0.245813	0.282684
2	0.387155	1.122749

And, for $h=0.25$ the values are,

t	y	$\frac{dy}{dt}$
0	1	$-1.1$
0.25	0.725	$-0.75219$
0.5	0.536593	$-0.45641$
0.75	0.422861	$-0.22728$
1	0.36603	$-0.0366$
1.25	0.356879	0.165057
1.5	0.398143	0.457865
1.75	0.51261	1.005997
2	0.764109	2.215916

Explanation of Solution

Given Information:

The initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$, with $y\left(0\right)=1$ in the interval $t\in \left(0,2\right)$ with step-size 0.5.

Formula used:

Solve an initial value problem of the form $\frac{dy}{dx}=f\left(x_i,y_i\right)$ using the iterative Euler method as,

$y_{i+1}=y_i+f\left(x_i,y_i\right)h$

Calculation:

From the initial condition $y\left(0\right)=1$, $y_0=1,t_0=0$, thus,

$\begin{array}{c}f\left(1,0\right)=\left(1\right){\left(0\right)}^2-1.1\left(1\right)\\ =-1.1\end{array}$

Let $h=0.5$, thus,

$\begin{array}{c}{y}_1=y_0+f\left(y_0,t_0\right)h\\ y\left(0.5\right)=y\left(0\right)+f\left(1,0\right)0.5\\ =1-1.1\left(0.5\right)\\ =0.45\end{array}$

Proceed further and use the following MATLAB code to implement Euler’s method and solve the differential equation.

% Q 1 (b) Euler's Method when h=0.5 and 0.25

clear;clc;

% Define the end-points of the interval [0,2]

a=0;b=2;

% Define the initial condition y(0)=1

y1(1)=1;y2(1)=1;

t1(1)=a;t2(1)=a;

dydt(1)=-1.1;

% Mention the step-size as 0.5 or 0.25

h1=0.5;h2=0.25

% Compute the value of the number of iterations corresponding to the step-size;

n1=(b-a)/h1;

n2=(b-a)/h2;

% Start the loop for the Euler's method

fori=1:n1

% Compute the value of the function to the right of

% the differential equation

dydt1(i+1)=(y1(i)*t1(i)*t1(i)-1.1*y1(i));

% Compute the value of the variable y

y1(i+1)=y1(i)+h1*dydt1(i+1);

% Compute the value of the variable t

t1(i+1)=t1(i)+h1;

end

% Display the results:

A1=[t1;y1;dydt1]'

fori=1:n2

% Compute the value of the function to the right of the % differentialequation

dydt2(i+1)=(y2(i)*t2(i)*t2(i)-1.1*y2(i));

% Compute the value of the variable y

y2(i+1)=y2(i)+h2*dydt2(i+1);

% Compute the value of the variable t

t2(i+1)=t2(i)+h2;

end

A2=[t2;y2;dydt2]'

t1=linspace(0,2);

exactsol=exp(((t1.^3)/3)-1.1*t1);

plot(t1, exactsol)

axis([0 2 0 2])

hold on

plot(A1(:,1), A1(:,2));hold on

plot(A2(:,1), A2(:,2))

xlabel('t')

legend('y(t)','h=0.5','h=0.25')

Execute the above code to obtain the solutions for $h=0.5$ tabulated as,

t	y	$\frac{dy}{dt}$
0	1	$-1.1$
0.5	0.45	$-0.3825$
1	0.25875	$-0.02588$
1.5	0.245813	0.282684
2	0.387155	1.122749

Now, the similar procedure can be followedfor the step size $h=0.25$.

The results thus obtained are tabulated as,

t	y	$\frac{dy}{dt}$
0	1	$-1.1$
0.25	0.725	$-0.75219$
0.5	0.536593	$-0.45641$
0.75	0.422861	$-0.22728$
1	0.36603	$-0.0366$
1.25	0.356879	0.165057
1.5	0.398143	0.457865
1.75	0.51261	1.005997
2	0.764109	2.215916

The results for the two-step-sizes are plotted along with the analytical solution $y\left(t\right)=e^{\frac{t^3}{3}-1.1t}+C$ as shown below,

It is inferred that the smaller step-size would give a better approximation to the solution.

To determine

To calculate: The solution of the initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$ with initial condition as $y\left(0\right)=1$ in the interval $t\in \left(0,2\right)$ and with the step-size of 0.5 using the mid-point method.

Answer to Problem 1P

Solution:

The solutions are tabulated as,

t	y	$\frac{dy}{dt}$
0	1	$-1.1$
0.5	0.623906	$-0.53032$
1	0.491862	$-0.04919$
1.5	0.602762	0.693176
2	1.364267	3.956374

Explanation of Solution

Given Information:

The initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$, with $y\left(0\right)=1$ in the interval $t\in \left(0,2\right)$ and with the step-size of 0.5.

Formula used:

Solve an initial value problem of the form $\frac{dy}{dx}=f\left(x_i,y_i\right)$ using the iterative mid-point method as,

$y_{i+1}=y_i+f\left(x_{i+0.5},y_{i+0.5}\right)h$

Here,

$y_{i+0.5}=y_i+f\left(x_i,y_i\right)\frac{h}{2}$

Calculation:

From the initial condition $y\left(0\right)=1$, $y_0=1,t_0=0$.

$\begin{array}{c}f\left(y_0,t_0\right)=f\left(1,0\right)\\ =\left(1\right){\left(0\right)}^2-1.1\left(1\right)\\ =-1.1\end{array}$

Let $h=0.5$. Thus,

$\begin{array}{c}{y}_{0+0.5}=y_0+f\left(x_0,y_0\right)\frac{h}{2}\\ =1-1.1\frac{\left(0.5\right)}{2}\\ =0.725\end{array}$

Now,

$\begin{array}{c}f\left(y_{0.5},t_{0.5}\right)=f\left(0.725,0.25\right)\\ =\left(0.725\right){\left(0.25\right)}^2-1.1\left(0.725\right)\\ =-0.75219\end{array}$

Proceed further and use the following MATLAB code to implement mid-point iterative scheme and solve the differential equation.

% Q 1 (c) Midpoint Method when h=0.5

clear;clc;

% Define the end-points of the interval [0,2]

a=0;b=2;

% Define the initial condition y(0)=1

y1(1)=1;

t1(1)=a;

% Mention the step-size as 0.5

h1=0.5;

% Compute the value of the number of iterations corresponding to the step-size;

n1=(b-a)/h1;

% Start the loop for the Euler's method

fori = 1: n1

% the value of the right hand side of the DE

b(i)=f1(t1(i), y1(i,:));

% Increment in t

t1(i+1) = t1(i) + h1;

% The formula for the Midpoint method

z = y1(i,:) + (h1/2) * b(i);

y1(i+1,:) = y1(i,:) + h1 * f1(t1(i)+(h1/2), z);

end

% Display the results:

A=[t1' y1]

t1=linspace(0,2);

exactsol=exp(((t1.^3)/3)-1.1*t1);

plot(t1, exactsol)

axis([0 2 0 2])

hold on

plot(A(:,1), A(:,2));

legend('y(t)','mid-point')

Execute the above code to obtain the solutions tabulated as,

t	Y	$\frac{dy}{dt}$
0	1	$-1.1$
0.5	0.623906	$-0.53032$
1	0.491862	$-0.04919$
1.5	0.602762	0.693176
2	1.364267	3.956374

The results for the are plotted along with the analytical solution $y\left(t\right)=e^{\frac{t^3}{3}-1.1t}+C$ as shown below,

Thus, it is inferred that the mid-point method gives a good approximation to the solution.

To determine

To calculate: The solution of the initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$ with the initial condition as $y\left(0\right)=1$ in the interval $t\in \left(0,2\right)$ with step-size of 0.5 using the fourth order RK method.

Answer to Problem 1P

Solution:

The solutions are tabulated as,

t	y	$k_1$	$k_2$	$k_3$	$k_4$
0	1	$-1.1$	$-0.7522$	$-0.8424$	$-0.4920$
0.5	0.6016	$-0.5113$	$-0.2546$	$-0.2891$	$-0.0457$
1	0.4645	$-0.0465$	0.2095	0.2391	0.6717
1.5	0.5914	0.6801	1.4953	1.8937	4.4609
2	1.5845	4.5949	10.8302	17.0071	51.9532

Explanation of Solution

Given Information:

The initial value problem $\frac{dy}{dt}=y{t}^2-1.1y$, with $y\left(0\right)=1$ in the interval $t\in \left(0,2\right)$ and with the step-sizes of 0.5 and 0.25.

Formula used:

Solve an initial value problem of the form $\frac{dy}{dx}=f\left(x_i,y_i\right)$ using the iterative fourth order RK method as,

$y_{i+1}=y_i+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)h$

In the above expression,

$\begin{array}{c}{k}_1=f\left(x_i,y_i\right)\\ k_2=f\left(x_i+\frac{h}{2},y_i+\frac{h}{2}{k}_1\right)\\ k_3=f\left(x_i+\frac{h}{2},y_i+\frac{h}{2}{k}_2\right)\\ k_4=f\left(x_i+h,y_i+k_{3}h\right)\end{array}$

Calculation:

From the initial condition $y\left(0\right)=1$, $y_0=1,t_0=0$.

$\begin{array}{c}{k}_1=f\left(y_0,t_0\right)\\ =f\left(1,0\right)\\ =\left(1\right){\left(0\right)}^2-1.1\left(1\right)\\ =-1.1\end{array}$

Let $h=0.5$. Thus,

$\begin{array}{c}{k}_2=f\left(t_0+\frac{0.5}{2},y_0+\left(\frac{0.5}{2}\right)k_1\right)\\ =f\left(0.25,0.725\right)\\ =-0.7522\end{array}$

And,

$\begin{array}{c}{k}_3=f\left(t_0+\frac{0.5}{2},y_0+\left(\frac{0.5}{2}\right)k_2\right)\\ =f\left(0.25,0.812\right)\\ =-0.8424\end{array}$

And,

$\begin{array}{c}{k}_4=f\left(t_0+0.5,y_0+0.5k_3\right)\\ =f\left(0.5,0.5788\right)\\ =-0.492\end{array}$

Therefore,

$\begin{array}{c}{y}_1=y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\left(0.5\right)\\ =1+\frac{1}{6}\left(-1.1+2\left(-0.7522\right)+2\left(-0.8424\right)+\left(-0.492\right)\right)\left(0.5\right)\\ =0.6016\end{array}$

Proceed further and use the following MATLAB code to implement RK method of order four, solve the differential equation.

% Q 1 (d) RK 4 Method when h=0.5

clear;clc;

% Define the end-points of the interval [0,2]

a=0;b=2;

% Define the initial condition y(0)=1

y1(1)=1;

t1(1)=a;

% Mention the step-size as 0.5

h1=0.5;

% Compute the value of the number of iterations corresponding to the step-size;

n1=(b-a)/h1;

halfh = h1 / 2;

h6 = h1/6;

% Start the process

fori = 1: n1+1

t1(i+1) = t1(i) + h1;

th2 = t1(i) + halfh;

% The values usually defined as k1, k2, k3 and k4

s1(i) = f1(t1(i), y1(i,:));

s2(i) = f1(th2, y1(i,:) + halfh * s1(i));

s3(i) = f1(th2, y1(i,:) + halfh * s2(i));

s4(i) = f1(t1(i+1), y1(i,:) + h1 * s3(i));

% The formula for the RK method of order 4

y1(i+1,:) = y1(i,:) + (s1(i) + s2(i)+s2(i) + s3(i)+s3(i) + s4(i)) * h6;

end;

t1(:, end:end)=[];

y1(end:end,:)=[];

% Display the values

A=[t1' y1 s1' s2' s3' s4']

% Plot the graph of the exact solution along with the approximations

t1=linspace(0,2);

exactsol=exp(((t1.^3)/3)-1.1*t1);

plot(t1, exactsol)

axis([0 2 0 2])

hold on

plot(A(:,1), A(:,2));

xlabel('t')

legend('y(t)','RK4')

In an another .m file, define the equation as,

functionfunc=f1(t, y)

func=y*t^2-1.1*y;

Execute the above code to obtain the solutions tabulated as,

t	y	$k_1$	$k_2$	$k_3$	$k_4$
0	1	$-1.1$	$-0.7522$	$-0.8424$	$-0.4920$
0.5	0.6016	$-0.5113$	$-0.2546$	$-0.2891$	$-0.0457$
1	0.4645	$-0.0465$	0.2095	0.2391	0.6717
1.5	0.5914	0.6801	1.4953	1.8937	4.4609
2	1.5845	4.5949	10.8302	17.0071	51.9532

The results for the are plotted along with the analytical solution $y\left(t\right)=e^{\frac{t^3}{3}-1.1t}+C$.

Hence, it is inferred that the RK method of order four gives the best approximation to the solution.