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(Solved): Given a first order IVP in normal form: \[ \left\{\begin{array}{l} y^{\prime}=f(x, y) \\ y\left(x_{ ...








Given a first order IVP in normal form:
\[
\left\{\begin{array}{l}
y^{\prime}=f(x, y) \\
y\left(x_{0}\right)=y_{0}
\end{array
Given a first order IVP in normal form: \[ \left\{\begin{array}{l} y^{\prime}=f(x, y) \\ y\left(x_{0}\right)=y_{0} \end{array}\right. \] we saw in class it is equivalent to solving the integral equation \[ y(x)=y_{0}+\int_{x_{0}}^{x} f(t, y(t)) d t \] This shows the solution we are looking for is a fixed point to the magic map \[ M[y]=y_{0}+\int_{x_{0}}^{x} f(t, y(t)) d t \] (Here \( M \) is a function whose input is a function, and output also a function!) This allows us to construct a sequence of iterates, starting with the initial guess \( y_{0}(x)=y_{0} \). (a constant function with value \( y_{0} \). the initial value), and \( y_{n+1}=M\left[y_{n}\right] \), that is. \[ y_{n+1}(x)=y_{0}+\int_{x_{0}}^{x} f\left(t, y_{n}(t)\right) d t \] If the hypothesis of Picard's theorem hold, then the sequence converges to a solution to the IVP locally! Let us try it out. Consider the IVP \[ \left\{\begin{array}{l} y^{\prime}=y^{2} \\ y(0)=1 \end{array}\right. \] Take the initial guess \( y_{0}(x)=1 \), and compute the iterates \( y_{1}(x) \) and \( y_{2}(x) \). If you do so correctly, \( y_{2}(x) \) should be a polynomial of degree 3 of the form \( A+B x+C x^{2}+D x^{3} \). Find \( A+B+C+D \) and enter your answer to 4 decimal places below. The next iterate \( y_{a}(x) \) is also a polynomial, What is the degree of the polynomial \( y_{3}(x) \) ?


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