Question

Use Euler's method with h=0.05 to find approximate values for the solution of the initial-value problem y'=2x^2+3y^2-2, y(2)=1 at x=0.1, 0.2, 0.3.

Answer 1

@SithsAndGiggles

Answer 2

Is \(h=0.05\) the step size?

Answer 3

Yes.

Answer 4

If so, all you need to do is construct a sort of table to keep track of what you're computing.

Clearly, \(f(x,y)=2x^2+3y^2-2\).
\[\begin{array}{c|c|c|c}
n&x_n&y_n&y_{n+1}=y_n+hf(x_n,y_n)\\
\hline
0&2&1&y_1=1+0.05(2(2)^2+3(1)^2-2)\\
1&2+h=2.05&y_1&y_2=y_1+0.05f(x_1,y_1)\\
\vdots&\vdots&\vdots&\vdots
\end{array}\]
I'm not sure what the question means by finding the approx solution value at \(x=0.1,0.2,0.3\)... Maybe it means when \(\Delta x=0.1,0.2,0.3\), i.e. when \(x_n=x_1+2h=2.1\), \(x_n=x_1+4h=2.2\), and \(x_n=x_1+6h=2.3\) ?

Answer 5

So I got \[y(2.05)\approx y _{1}=y _{0}+f(2, 1)(0.05)=1.45\]

Answer 6

@hartnn @amistre64 What do I do next?

Answer 7

@aaronq @dan815 @robtobey