Edit: look at the picture at the bottom first. I'm meddling with exercise 2.
Please answer avoiding ODEs, since this course comes before ODEs in my school.
So far, my asnwer to ex. 1 is the following:
I have a two variable scalar function $f$ such that $f(x_0,y)=x_0$ for every $y$, i.e. $x_0$ is a fixed point of $f$ for any $y$.
I have proved that there is a function $f_1(x,y)$ such that $f(x,y)=x_0 + f_1(x,y)(x-x_0),$ in the following way:
$$f(x,y)-x_0=f(x,y)-f(x_0,y)= \int_{x_0}^x \frac{d}{dx} f(z,y) dz,$$
and applying the change of variable $z = x_0 + t(x-x_0)$, which gives $t=0$ when $z = x_0$ and $t=1$ for $z= x$; and $dz= (x-x_0)dt$, we get:
$$\int_{x_0}^x \frac{d}{dx} f(z,y) dz = \int_0^1 \frac{d}{dx} f(x_0 +t(x-x_0),y) dt (x-x_0).$$
Then the demanded function is $f_1(x,y) = \int_0^1 \frac{d}{dx} f(x_0 +t(x-x_0),y) dt$.
Now, I'm asked to take the first and second partial derivatives with respect to $x$ and $y$ of $f_1(x,y)$, using (some of) the sufficient and/or non-degeneracy conditions for a transcritical bifurcation: $$\frac{\partial}{\partial x}f(x_0,y)=1,$$ $$\frac{\partial}{\partial y}f(x_0,y)=0,$$ $$\frac{\partial^2}{\partial x^2}f(x_0,y)\neq1,$$ $$\frac{\partial^2}{\partial x^2}f(x_0,y) \frac{\partial^2}{\partial x^y}f(x_0,y) - \Big(\frac{\partial^2}{\partial x y}f(x_0,y)\Big)^2 < 0.$$
I'm completely stuck at this point, I visited and revisited the fisrt and second fundamental theorem of calculus but I still can't fugire it out. Can you help me?
Thanks in advance.
Edit: this is the exercise in which this problem appears:
