# Solving a differential equation

• JulieK

#### JulieK

I have the following differential equation

$\frac{\partial}{\partial t}\left(\frac{a}{X}\right)+\frac{X}{b}\frac{ \partial Y}{\partial t}+\frac{c}{X}=0$

where $a$, $b$ and $c$ are constants and $X$ is a function of
$t$. I want to solve it for $Y$ analytically (if possible) or numerically.

First, that's not really a partial differential equation because the only differentiation is with respect to the single variable, t. If X(t) is a known function of t, then $\partial/\partial ta/X$ is also a known function of t- call it X'(t). Then your equation can be written
$$\frac{X}{b}\frac{dY}{dt}= -X'- C/X$$

$$\frac{dY}{dt}= (-X'- C/X)\frac{b}{X}= -\frac{X'X- C}{X^2}$$
and you solve for Y by integrating.b

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If X(t) is a known function of t I would solve it easily. Unfortunately this is not the case.

Then all you can do is write
$$Y(t)= -\int\frac{XX'- C}{X^2}dt$$

If X(t) is a known function of t I would solve it easily. Unfortunately this is not the case.

Two unknown functions and one equation only is not enough. You need two equations.

Many thanks to you all!

To close the gap, I obtained a second relation

$Y=\frac{d}{O}\left(X^{1/2}-O^{1/2}\right)$

where $d$ is a constant and $O$ is a known function of $t$ with
a closed analytical form.

First, that's not really a partial differential equation because the only differentiation is with respect to the single variable, t. If X(t) is a known function of t, then $\partial/\partial ta/X$ is also a known function of t- call it X'(t). Then your equation can be written
$$\frac{X}{b}\frac{dY}{dt}= -X'- C/X$$

$$\frac{dY}{dt}= (-X'- C/X)\frac{b}{X}= -\frac{X'X- C}{X^2}$$
and you solve for Y by integrating.b

I would be inclined to call ##\frac{d}{dt}(a/X(t))## something other than ##X'(t)##, which could be confused with ##dX/dt##. :tongue2:

Many thanks to you all!

To close the gap, I obtained a second relation

$Y=\frac{d}{O}\left(X^{1/2}-O^{1/2}\right)$

where $d$ is a constant and $O$ is a known function of $t$ with
a closed analytical form.
Take the derivative w.r.t. t, then you can substitute for $\frac{dY}{dt}$ in HallsofIvy's equation in Post #2. Now you have a differential equation in just one unknown function, X(t).

p.s. "d" is not the best name for a quantity in anything having to do with calculus