I am solving this equation;
DSolve[{y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}, y[t], t] NDSolve[{y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}, y, {t, 0, 1.3}] I have solved it numerically and analytically and I have plotted both results.
I need to check my solutions by substituting them back into the original equation; I am not sure how to do this, specifically for the numerical solution.
An other way.
dsol = DSolve[{y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}, y, t] ndsol = NDSolve[{y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}, y, {t, 0, 1.3}, InterpolationOrder -> All] {eq, ic} = {y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}; Plot[(y[t] /. First@dsol) - (y[t] /. First@ndsol), {t, 0, 1.3}, PlotRange -> All] ic /. First@ndsol // Simplify (* True *) Plot[eq[[1]] - eq[[2]] /. First@ndsol, {t, 0, 1.3}, PlotRange -> All] Something like this?
sol1 = y[t] /. DSolve[{y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}, y[t], t] sol2 = y[t] /. NDSolve[{y'[t] == (t^2 y[t]^2)/(1 + 8 t^4), y[0] == 1/2}, y, {t, 0, 1.3}] Plot[{Abs[sol1 - sol2]}, {t, 0, 1.3}] 
ycould be huge while the error in the ODE should still be close to zero. Checking between the steps will add a significant interpolation error, unless you use the optionInterpolationOrder -> AllinNDSolve. $\endgroup$