Use Euler's method with h=0.05 to find approximate values?

[Math10]

Homework Statement

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.

Homework Equations

None.

The Attempt at a Solution

Here's my work:

y'=2x^2+3y^2-2, y(2)=1
f(x, y)=2x^2+3y^2-2, x₀=2, y₀=1
y(0.1)=y₁=y₀+f(2, 1)(0.05)=1.45
y(0.2)=y₂=y₁+f(2.1, 1.45)(0.05)=2.10638
y(0.3)=y₃=y₂+f(2.3, 2.10638)(0.05)=3.20091
The answers I got are 1.45, 2.10638 and 3.20091. But are my answers right? Please check my answer and correct me if I'm wrong.

 

[HallsofIvy]

[No, those are not correct.. If dy/dx= 2x^2+3y^2-2, then dy= (2x^2+3y^2-2)dx
[itex]y_n(x_n)+ f(x_n, y_n)(dx)[/itex] is [itex]y(x_{n+1)}= y(x_n+ dx)[/itex].
[itex]y_0+ f(2, 1)(0.05)[/itex] is [itex]y(0+ 0.05)= y(0.05)[/itex] NOT [itex]y(0.1)[/itex].

 

[vela]

[No, those are not correct. If dy/dx= 2x^2+3y^2-2, then dy= (2x^2+3y^2-2)dx

[itex]y_n(x_n) + f(x_n, y_n)(dx)[/itex] is [itex]y(x_{n+1}) = y(x_n+ dx)[/itex].

[itex]y_0+ f(2, 1)(0.05)[/itex] is [itex]y(0+ 0.05)= y(0.05)[/itex] NOT [itex]y(0.1)[/itex].

Since you're evaluating f at x=2, that should be y(2.05) = y(2)+f(2,1)(0.05).

 

[Math10]

So how do I start the problem?
y(2.05)=y(2)+f(2, 1)(0.05)=1.45
But where did y(2.05) come from?

 

[vela]

What's the basic idea of Euler's method? You should understand that before trying to do this problem. It'll make the calculations make much more sense. Right now it seems like you're just plugging in numbers without knowing why.

 

[Math10]

To find approximate values for the solution of the initial-value problem at the given points.

 

[vela]

That's so general it describes every method for numerically solving a differential equation. What specifically is the idea behind Euler's method?