"Given dy/dx=x+2y and y=0 when x=1, use Euler's method with increments of deltax=.2 to approximate when x=1.6. Show all work"
4 answers
I would also like to see the answer to this one!
so, just apply the method three times.
y(x+h) = y(x) + y'(x)*h
y(1.2) = y(1) + (1+2*0)(0.2) = 0.6
y(1.4) = y(1.2) + (1.2+2*0.6)(0.2) = 1.08
and so on
y(x+h) = y(x) + y'(x)*h
y(1.2) = y(1) + (1+2*0)(0.2) = 0.6
y(1.4) = y(1.2) + (1.2+2*0.6)(0.2) = 1.08
and so on
Can you please complete the question? I am still slightly confused.
I gave you the formula and two examples.
You need to read up on Euler's method. It just relies on the fact that given the tangent line at (x,y), ∆y/∆x ≈ dy/dx
You need to read up on Euler's method. It just relies on the fact that given the tangent line at (x,y), ∆y/∆x ≈ dy/dx
1 answer