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Use the two second-order multi-step methods

$$ω_{i+1} = ω_i + \frac{h}{2}(3f_i − f_{i−1})$$ and $$ω_{i+1} = ω_i +\frac{h}{2}(f_{i+1}+ f_i)$$

as a predictor-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.

I do not understand how to use these methods to approximate $y(0.3)$. Moreover I am not sure how Euler's method fits into this question. Could someone clarify this question please?

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The $$ \omega_{i+1} = \omega_i + \frac{h}{2}(3f(t_i,\omega_i) - f(t_{i-1},\omega_{i-1})) $$ is an explicit two-step (using three points $i-1, i$ and $i+1$) method. In contrast, the second one $$ \omega_{i+1} = \omega_i + \frac{h}{2}(f(t_{i+1},\omega_{i+1}) + f(t_{i},\omega_i)) $$ 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 $$ \omega_{i+1} - \frac{h}{2}f(t_{i-1},\omega_{i+1}) = \omega_{i} + \frac{h}{2}f(t_{i},\omega_{i}). $$ Instead of solving that equation you're advised to use a predictor-corrector scheme, i.e. use $\omega_{i+1}$, obtained from the first scheme and plug it into $f(\omega_{i+1})$ term of the second one. I'll rewrite it using tilde notation ($\tilde \omega_{i+1}$ is a predicted value, while $\omega_{i+1}$ is a corrected one): $$ \text{Predictor: }\tilde{\omega}_{i+1} = \omega_i + \frac{h}{2}(3f(t_{i},\omega_i) - f(t_{i+1},\omega_{i-1}))\\ \text{Corrector: }\omega_{i+1} = \omega_i + \frac{h}{2}(f(t_{i+1},\tilde{\omega}_{i+1}) + f(t_{i},\omega_i)). $$ Now the evaluation is really straitforward, you just take $\omega_{i-1}, \omega_{i}$, compute $\tilde{\omega}_{i+1}$ 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 predictor-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: $$ \omega_1 = \omega_0 + h f(t_0,\omega_0). $$

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