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