# New Approximation Methods Based on Fuzzy Transform for Solving SODEs: II
## 4 New Fuzzy Numerical Methods for Solving SODEs
### Numerical Scheme III: Modified 3-Step Adams Moulton Method Based on FzT and NIM for SODEs  
In this subsection, we improve 3-step Adams Moulton method using FzT and NIM for solving SODEs (16). The modified 3-step Adams Moulton method proposed by [16] for solving Cauchy problems. From this idea, we can propose to approximate the solution of (16) by NIM and FzT components, \(X_{k}\ \left(Y_{k}\right)\), \(k=2,\ldots,n-1\), of the unknown function \(x(t)\ \left(y(t)\right)\) with respect to choose some of the \(h\)-uniform generalized fuzzy partition (see Definition 4), \(B_{1},\ldots,B_{n}\), of interval \([a,b]\) with parameter \(h=\left(b-a\right)/\left(n-1\right)\), \(n\geq 2\). Let \(X_{1}=x_{1},\)\(Y_{1}=y_{1},\)\(X_{2}=x_{2},\)\(Y_{2}=y_{2},\)\(X_{3}=x_{3},\) and \(Y_{3}=y_{3}\) if possible; otherwise, we can compute FzT components \(X_{2},\ Y_{2},\ X_{3}\) and \(Y_{3}\) from the numerical Scheme I. Now, we apply the F -transform and NIM to the SODEs (16) and obtain the following numerical Scheme III for \(k=3,\ldots,n-1\):  
According to steps of deriving Equation (27) and then steps of NIM in previous Subsection 4.2, we get the four term approximation of the NIM as follows.  
\[\begin{array}{rcl}u_{x0}&=&X_{k}+\frac{h}{24}\left(19F_{k}-5F_{k-1}+F_{k-2} \right),\\ u_{x1}&=&N\left(u_{x0}\right),\\ u_{x2}&=&N\left(u_{x0}+u_{x1}\right)-N\left(u_{x0}\right),\\ u_{x3}&=&N\left(u_{x0}+u_{x1}+u_{x2}\right)-N\left(u_{x0}+u_{x1}\right).\\ \end{array}\left|\begin{array}{rcl}u_{y0}&=&Y_{k}+\frac{h}{24}\left(19G_{k}-5 G_{k-1}+G_{k-2}\right),\\ u_{y1}&=&N\left(u_{y0}\right),\\ u_{y2}&=&N\left(u_{y0}+u_{y1}\right)-N\left(u_{y0}\right),\\ u_{y3}&=&N\left(u_{y0}+u_{y1}+u_{y2}\right)-N\left(u_{y0}+u_{y1}\right).\\ \end{array}\right\}\end{array}\]  
Hence, the four term approximate solution is  
\[u_{x}=u_{x0}+u_{x1}+u_{x2}+u_{x3}=u_{x0}+N\left(u_{x0}+N\left(u_{x0}+u_{x1} \right)\right)\]  
and  
\[u_{y}=u_{y0}+N\left(u_{y0}+N\left(u_{y0}+u_{y1}\right)\right),\]  
which leads to the following formulas.  
\[\left.\begin{array}{rcl}X_{k+1}^{*}&=&X_{k}+\frac{h}{24}\left(19F_{k}-5F_{k- 1}+F_{k-2}\right),\\ X_{k+1}^{**}&=&X_{k+1}^{*}+\frac{9h}{24}F_{k+1}^{*},\\ X_{k+1}^{***}&=&X_{k+1}^{*}+\frac{9h}{24}C_{k+1}^{*},\\ X_{k+1}^{***}&=&Y_{k+1}^{*}+\frac{9h}{24}C_{k+1}^{**},\\ X_{k+1}&=&X_{k},\\ &+&\frac{h}{24}\left(19F_{k}-5F_{k-1}+F_{k-2}+9F_{k+1}^{***}\right),\end{array} \right\} \tag{29}\]  
where  
\[\begin{array}{rcl}F_{k-2}&=&\frac{\int_{a}^{b}f(t,X_{k-2},Y_{k-2})B_{k-2}(t) dt}{\int_{a}^{b}B_{k-2}(t)dt},\\ F_{k-1}&=&\frac{\int_{a}^{b}f(t,X_{k-1},Y_{k-1})B_{k-1}(t)dt}{\int_{a}^{b}B_{k- 1}(t)dt},\\ F_{k}&=&\frac{\int_{a}^{b}f(t,X_{k},Y_{k})B_{k}(t)dt}{\int_{a}^{b}B_{k}(t)dt}, \\ F_{k+1}^{*}&=&\frac{\int_{a}^{b}f(t,X_{k+1}^{*},Y_{k+1}^{*})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt},\\ F_{k+1}^{**}&=&\frac{\int_{a}^{b}f(t,X_{k+1}^{**},Y_{k+1}^{**})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt},\\ F_{k+1}^{***}&=&\frac{\int_{a}^{b}f(t,X_{k+1}^{**},Y_{k+1}^{**})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt},\\ F_{k+1}^{***}&=&\frac{\int_{a}^{b}f(t,X_{k+1}^{***},Y_{k+1}^{**})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt},\\ F_{k+1}^{***}&=&\frac{\int_{a}^{b}f(t,X_{k+1}^{***},Y_{k+1}^{**})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt},\\ F_{k+1}^{***}&=&\frac{\int_{a}^{b}f(t,X_{k+1}^{***},Y_{k+1}^{***})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt},\\ F_{k+1}^{***}&=&\frac{\int_{a}^{b}g(t,X_{k+1}^{***},Y_{k+1}^{***})B_{k+1}(t)dt}{ \int_{a}^{b}B_{k+1}(t)dt}.\\ \end{array}\]  
In the sequel, the inverse FZT (25) approximates the solution \(x(t)\ (y(t))\) of the problem (16).