Question

Use Euler's method with h=0.1 to find approximate values for the solution of the initial-value problem y'+2y=(x^3)(e^-2x), y(0)=1 at x=0.1, 0.2, 0.3.

Answer 1

\[y'=-2y+x ^{3}e ^{-2x}, y(0)=1\]

Answer 2

@SithsAndGiggles

Answer 3

I don't quite remember the numerical methods for solving DEs... give me a while to catch up on the material?

Answer 4

\[f(x, y)=-2y+x ^{3}e ^{-2x}, x _{0}=0,y _{0}=1\]

Answer 5

\[y _{1}=y _{0}+hf(x _{0}, y _{0})=1+(0.1)f(0, 1)=0.8\]

Answer 6

Never mind. I got it. Thanks for the help.