How Can I Solve This Differential Equation for Y?

[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.

 

[HallsofIvy]

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

 

[JulieK]

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

 

[HallsofIvy]

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

 

[JJacquelin]

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.

 

[JulieK]

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.

 

[Mute]

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 $dX/dt$.

 

[Redbelly98]

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