What does the notation $f_y$ and $f_{y'}$ mean for this differential equation

Question

Consider the Differential Eqn $$y''(t) + y(t) = \tan(t).$$ So an equivalent form of this Differential Equation is $$y'' = f(t, y(t), y'(t))$$ where $$f(t, y(t), y'(t)) = \tan(t)-y(t).$$ My numerical analysis textbook has an algorithm that has the following notation.

$$f_y(t, y(t), y'(t)) \text{ and } f_{y'}(t, y(t), y'(t))$$

I assume that these are the partial derivatives with respect to $y$ and then with respect to $y'$. However I am not exactly sure.

So, my question. With respect to the original Differential Equation, what is $f_y$ and $f_{y'}$?

Would $f_y = -1$ and $f_{y'} = 0$?

Answer 1

$~f(t, y(t), y'(t))~$ is a twice continuously differentiable function and these two partial derivatives $~f_y~$ and $~f_{y'}~$ are the derivatives of $~f(t, y(t), y'(t))~$ with respect to $~y~$ and with respect to $~y'~$ respectively.

The value of $~f_y~$ and $~f_{y'}~$ depend only on the function $~f(t, y(t), y'(t))~$.

• Let us discuss the matter by giving a small example.

Example, let $~f(t, y(t), y'(t))=y'^2+12ty~$.

Then $~f_y(t, y(t), y'(t))=\dfrac{d}{dy}f(t, y(t), y'(t))=12t~~$ and

$~~~~~~~~~ f_{y'}(t, y(t), y'(t))=\dfrac{d}{dy'}f(t, y(t), y'(t))=2y'~$.

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Note: In your case, $~f(t, y(t), y'(t)) = \tan(t)-y(t)~$

So $~f_y(t, y(t), y'(t))=\dfrac{d}{dy}f(t, y(t), y'(t))=-1~~~$ and

$~~~~~ f_{y'}(t, y(t), y'(t))=\dfrac{d}{dy'}f(t, y(t), y'(t))=0~$.