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Consider the IVP $x'(t) = 1 + (x(t))^2,\quad x(0) = 0.$ Backward Euler method is applied to this problem the numerical solution $x_1$ at time $t_1 = h$.Then I found $x_1= \frac{1+ (1- 4h^2)^{1/2}}{2h}$ but solution say $X_1=\frac{2h}{1 +(1-4h^2)^{1/2}}$

Where is the mistake? $\begin{align*} X_{n+1}= X_n +hX'_{n+1} &\implies X_{n+1}= X_n +h( 1 +(X_{n+1})^2)\\ &\implies x(h)= X(0)+h(1+X(h)^2)\\ &\implies hx(h)^2 -X(h)+h\\ &\implies X_1= \frac{1+(1-4hh)^{1/2}}{ 2h} &\text{(by root finding formula)} \end{align*}$

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    It seems you have a few problems with the backward Euler method. Do you have a teacher you can go to?2012-04-22

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Small problems: you go back and forth between $x$ and $X$ and $X_1$.

Also, I think you left out the "$=0$" on the next-to-last line.

Bigger problem: a quadratic equation (generally) has two solutions. You have left out the $\pm$. Then you might meditate on whether ${2h\over1+(1-4h^2)^{1/2}}{\rm\quad and\quad }{1-(1-4h^2)^{1/2}\over2h}$ are all that different.