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In the implicit Euler method, we estimate the value of the function at a point using the slope at that point (x-value), which first seems absurd. As how can we know the slope at a point if we don't know the point itself?

But I came to realize that we solve the implicit equation of $y_{n+1}$ $$y_{n+1}=y_n + h*f(x_{n+1},y_{n+1})$$ let $f(x,y)= x+y$

then, finally solving the above equation for $y_{n+1}$ becomes $$y_{n+1}=\frac{y_n+x_{n+1}}{(1-h)}$$

But how will this work when the variables are not separable in the implicit equation? For example, I can think of maybe $f=xy^2$. Here again, how can we make $y_{n+1}$ the subject of the equation?

Does this become a limitation for the backward Euler method, when we can't separate out $y_{n+1}$? Also, explain the very first statement, which doesn't accord with me on an intuitive level.

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    $\begingroup$ You are correct, any implicit method, which backward Euler is the simplest example of, applied to a non-linear ODE may not be solvable and would require additional techniques, usually some form of linearization, e.g. Newton's method. The best intuition to me for the backward Euler method's approach comes from the quadrature rule form (take a look at github.com/mandli/intro-numerical-methods/blob/master/…) for an explanation). $\endgroup$ Commented Feb 28 at 13:51

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As you noted, the implicit Euler method requires you to solve the equation $$ y_{n+1} = y_n + h f(x_{n+1},y_{n+1}) $$ for $y_{n+1}$. If $f$ has a simple enough formula, then you can separate $y_{n+1}$ to one side of the equation and express it in terms of $x_{n+1}$ and $y_n$. In general, if $f$ is a complicated formula, you can't separate $y_{n+1}$ to one side, and in that case you are simply stuck with a nonlinear equation for $y_{n+1}$ that you need to solve via a method such as Newton's method.

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This is the property of implicit methods in general, not just the simple backward Euler. You normally have to solve a set of algebraic equations at each time step and usually iteratively. That makes the process for each time step a) more complicated, b) slower to compute. However, you can often use longer time steps due to the better stability of implicit methods.

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