# Use the two second-order multi-step methods omega_(i+1)=omega_i+h/2(3f_i−f_(i−1)) and omega_(i+1)=omega_i+h/2(f_i+1+f_i) as a pblackictor-corrector method to compute an approximation to y(0.3), with stepsize h=0.1, for the IVP; y′(t)=3ty,y(0)=−1. Use Euler’s method to start.

Use the two second-order multi-step methods
${\omega }_{i+1}={\omega }_{i}+\frac{h}{2}\left(3{f}_{i}-{f}_{i-1}\right)$
and
${\omega }_{i+1}={\omega }_{i}+\frac{h}{2}\left({f}_{i+1}+{f}_{i}\right)$
as a pblackictor-corrector method to compute an approximation to $y\left(0.3\right)$, with stepsize $h=0.1$, for the IVP;
${y}^{\prime }\left(t\right)=3ty,y\left(0\right)=-1.$
Use Euler’s method to start.

I do not understand how to use these methods to approximate $y\left(0.3\right)$. Moreover I am not sure how Euler's method fits into this question. Could someone clarify this question please?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

The
${\omega }_{i+1}={\omega }_{i}+\frac{h}{2}\left(3f\left({t}_{i},{\omega }_{i}\right)-f\left({t}_{i-1},{\omega }_{i-1}\right)\right)$
is an explicit two-step (using three points $i-1$,$i$ and $i+1$) method. In contrast, the second one
${\omega }_{i+1}={\omega }_{i}+\frac{h}{2}\left(f\left({t}_{i+1},{\omega }_{i+1}\right)+f\left({t}_{i},{\omega }_{i}\right)\right)$
is an implicit one-step (using two points $i$ and $i+1$) method. It is implicit since you can not easily solve it for ${\omega }_{i+1}$, you have an equation for it
${\omega }_{i+1}-\frac{h}{2}f\left({t}_{i-1},{\omega }_{i+1}\right)={\omega }_{i}+\frac{h}{2}f\left({t}_{i},{\omega }_{i}\right).$
Instead of solving that equation you're advised to use a pblackictor-corrector scheme, i.e. use ${\omega }_{i+1}$, obtained from the first scheme and plug it into $f\left({\omega }_{i+1}\right)$ term of the second one. I'll rewrite it using tilde notation (${\stackrel{~}{\omega }}_{i+1}$ is a pblackicted value, while ${\omega }_{i+1}$ is a corrected one):

Now the evaluation is really straitforward, you just take ${\omega }_{i-1},{\omega }_{i}$, compute ${\stackrel{~}{\omega }}_{i+1}$ and finally ${\omega }_{i+1}$. The only problem is that you cannot start with that. The only initial value you have is ${\omega }_{0}=-1$ which is not sufficient to compute ${\omega }_{1}$ using pblackictor-corrector pair (it would need also ${\omega }_{-1}$). To overcome this problem you can use some other method to compute ${\omega }_{1}$, explicit Euler for example:
${\omega }_{1}={\omega }_{0}+hf\left({t}_{0},{\omega }_{0}\right).$