number analysis: Given y' = ?ym y(0) = 1, verify y(t) = e^(?

[mathmathmath9]

Given y' = ?ym y(0) = 1

verify the solution to this equation is y(t) = e^(?t). by solving this equation numerically to get y(1).

consider these approximation solution to this equations:
Method 1:
yn+1 = yn + hf(yn+1,tn+1)
Method 2:
yn+1= yn + (h/2)(f(yn,tn) + f(yn+1,tn+1))

1.computer the truncation error for both methods. which is moe accurate?
2.substitute the value of f(y,t) = ?y and solve for yn+1 in both methods.
3.obtain an expression for yn+1 as a function of y0 for both methods
4.using the expressions obtained in the previous part, compute limit, as n-> infiniti, of yn
5.do the methods converge?

 

[galactus]

What is m?. An arbitrary constant?.

\(\displaystyle \frac{dy}{dt}={\lambda}ym, \;\ y(0)=1\)

Separate variables:

\(\displaystyle \frac{dy}{y}={\lambda}mdt\)

Integrate:

\(\displaystyle ln(y)={\lambda}mt+C\)

\(\displaystyle y=e^{{\lambda}mt+C}=C_{1}e^{{\lambda}mt}\)

Using I.C.:

\(\displaystyle C_{1}e^{{\lambda}m(0)}=1\)

\(\displaystyle C_{1}=1\)

Therefore, we have \(\displaystyle y=e^{{\lambda}mt\)