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**HallsofIvy** It **isn't** equal to $\displaystyle t^2- 1$. Where did you get that? You must be looking at an example that has nothing to do with this given problem. f(t, Yn(t)) is simply f applied to t and whatever you have just calculated Yn(t) to be!

In this case $\displaystyle dy/dx= x^2y$. Since we are given that y(0)= 1, a reasonable starting approximation for y is the constant y= 1; Then we have $\displaystyle dy/dx= x^2(1)$ and, integrating $\displaystyle y= (1/3)x^3+ C$. In order to have y(0)= 1, we must have y(0)= (1/3)0+ C= 1 so C= 1 and $\displaystyle y= (1/3)x^3+ 1$. Replacing the y in the differential equation with that, $\displaystyle dy/dx= x^2((1/3)x^3+ 1)= (1/3)x^5+ x^2$ and integrating, $\displaystyle y= (1/18)x^6+ (1/3)x^3+ C$. Again, y(0)= 1 gives C= 1. Take $\displaystyle dy/dx= f(x,y(x))= x^2((1/18)x^3+ (1/3)x^3+ 1)$ and integrate again.