Solving 4th Order Runge-Kutta w/ dy/dt = y, y(0)=1

[kmeado07]

Homework Statement

Use the fourth order Runge-Kutta formula to advance the differential equation:
dy/dt = y with y(0)=1 forward one step h. That is find y(h).

Homework Equations

The Attempt at a Solution

The Runge-Kutta formula is:
x(i+1)=x(i)+h/6 [k(1)+2k(2)+2k(3)+k(4)]

where,
k(1)=f[x(i),t(i)]
k(2)=f[x(i)+1/2 hk(1), t(i)+1/2 h]
k(3)=f[x(i)+1/2 hk(2), t(i)+1/2h]
k(4)=f[x(i)+hk(3),t(i)+h]

I have no idea how to continue though, so any help would be great! thanks

 

[bsodmike]

A first-order ODE is given as,

dy/dx = f(x,y)

this is the defintion. So when one states, f(x_i,y_i), this simply means, evaluate the first-order derivative at x_i and y_i.

This is how the method breaks down:

The general form of the RK4 algorithm is given as,<br /> \begin{equation}\begin{split} <br /> k_{1}&amp;=f(x_{i},y_{i})\\<br /> k_{2}&amp;=f(x_{i}+\dfrac{1}{2}h, y_{i}+\dfrac{1}{2}k_{1}h)\\<br /> k_{3}&amp;=f(x_{i}+\dfrac{1}{2}h,y_{i}+\dfrac{1}{2}k_{2}h)\\<br /> k_{4}&amp;=f(x_{i}+h,y_{i}+k_{3}h)\\<br /> y_{i+1}&amp;=y_{i}+\dfrac{1}{6}\cdot(k_{1}+2k_{2}+2k_{3}+k_{4})\cdot h<br /> \label{eq:}<br /> \end{split}\end{equation}

Simplified as eight equations,<br /> \begin{equation}\begin{split} <br /> k_1&amp;=f(x_i,y_i)\\<br /> z_1&amp;=y_i+\dfrac{1}{2}k_{1}h\\<br /> k_2&amp;=f(x_{i}+\dfrac{1}{2}h, z_1)\\<br /> z_2&amp;=y_i+\dfrac{1}{2}k_{2}h\\<br /> k_3&amp;=f(x_{i}+\dfrac{1}{2}h, z_2)\\<br /> z_3&amp;=y_i+k_{3}h\\<br /> k_4&amp;=f(x_{i}+h, z_3)\\<br /> y_{i+1}&amp;=y_i+\left(\dfrac{h\cdot(k_1+2(k_2+k_3)+ k_4)}{6}\right)<br /> \label{eq:}<br /> \end{split}\end{equation}<br />

Say for example you are given, y&#039;=(-y)ln(y),\;\;y(0)=0.5.

Hence,
f(x_0,y_0)=(-0.5)\cdot ln(0.5)

In this example, it only varied in terms of y. However, you could be faced with an ODE such as y&#039;=4e^{0.8x}-0.5y, in which case you need to take care how you substitute in the 'x' value. For the x_i+\dfrac{1}{2}h bits, you need to evaluate it at x_{i+1}=x_i+\dfrac{1}{2}h.

For example, if x_0=0 and h=0.5, then x_1=0+\dfrac{1}{2}(0.5)=0.25.

Hope this helps...

 

[kmeado07]

thankyou very much for that...i understand all of what you have said.

So i found for my equation that f[t(0),y(0)] = 1, since y'=y and y(0)=1

Im still slightly confused as to how to find y(h)?

 

[kmeado07]

How do i find what k(1) - k(4) is? Do i substitute anything in for h?

Also what are the value for t(i) and y(i)?

 

[bsodmike]

You shouldn't jump to using RK4 if you don't understand the basics. Try using the Euler method (it's basically a first-order RK). You also need to know the stepsize 'h'. This is given with questions,

y_{i+1}=y_i+hf(x_i,y_i)

As for your question, y'=y, you can ignore the x or 'time' component as it were, as it only varies in y.

The first RK4 application would be, note y_0=y(0)=1,

k_1 = f(0, 0) =1
k_2 = y_0+0.5*k_1*h
k_3 = y_0+0.5*k_2*h
k_4 = y_0+k_3*h

Hence,
y_1=y_0+\left(\dfrac{h*(k_1+2(k_2+k_3)+k_4)}{6}\right)

For the next iteration, use y_1 to start off :)