Fixing the initial value, how does a change in parameters affect the solution to a differential equation?

145 Views Asked by At

Take the following differential equation with $a\in \mathbf{R}$, and $g:\mathbf{R}\rightarrow\mathbf{R}^{++}$ with $g$ continuously differentiable.

$$\dot{y}=\frac{a}{g(y(t))}$$

and initial value $y(0)=0$.

For a given $t>0$ and keeping the initial value unchanged, how can I calculate how $a$ affects $y(t)$?


I can write $y(t) = y(0) + \int_0^t \dot{y}(t)dt $. Then taking the derivative

$$\dfrac{\partial y(t)}{\partial a}=\int_0^t \dfrac{\partial\dot{y}(t)}{\partial a}dt=\int_0^t \frac{1}{g(y(t))}-a\dfrac{g'(y(t))}{g(y(t))^2}\dfrac{\partial y(t)}{\partial a}dt$$

  1. Is this the correct approach?
  2. If yes, what would I do next with this integral equation? How can I determine at least the sign of $\dfrac{\partial{y(t)}}{\partial a}$?

To clarify: I am not looking for a solution for a specific $g$, but rather for what the correct steps are to solve this for a general $g$.

1

There are 1 best solutions below

1
On

This is a separable differential equation

It becomes $\ln (y(t))=at +c$

When $t= 0$ then $y(0)=0$

You get $c=0$

The equation becomes $e^{at}=y(t)$

Now you can see how $a$ affects $y(t)$

And about your method is just writing same thing on L.H.S and R.H.S