First-order ODE

Question

Please look at the equation

$$ \frac{d x }{dt } = \frac{c}{x} - x + h(t) $$

where $c \geq 0$ is a constant; the initial condition is given at time $t=0$ (say $x = x_0$ at $t=0$); and $h(t)$ is a function defined for all $t\geq 0$.

For $c=0$ the solution is

$$ x = x_0e^{-t} + \int_0^t e^{-(t-\tau)} h(\tau) d \tau $$

For $h=0$ the solution is

$$ x = \sqrt{c + (x_0^2 - c) e^{-2 t} } $$

What would the solution be for both $c$ and $h$ are different from zero?

Answer 1

Ｓuppose $x_0>0$．Note that $$ x'=\frac{c}{x}-x $$ has the GS $$ x(t)=\sqrt{c+(C-c)e^{-2t}} $$ Let $x(t)=\sqrt{c+(u(t)-c)e^{-2t}}$ be the solution of $$ \frac{d x }{dt } = \frac{c}{x} - x + h(t). $$ Then $u(t)$ satisfies $$ e^{-t}u'(t)=2h(t)\sqrt{c+(u(t)-c)e^{-2t}}$$ which can't be solved explicitly.

Answer 2

Hint:

$\dfrac{dx}{dt}=\dfrac{c}{x}-x+h(t)$

$x\dfrac{dx}{dt}=-x^2+h(t)x+c$

This belongs to an Abel equation of the second kind.

Let $x=e^{-t}u$ ,

Then $\dfrac{dx}{dt}=e^{-t}\dfrac{du}{dt}-e^{-t}u$

$\therefore e^{-t}u\left(e^{-t}\dfrac{du}{dt}-e^{-t}u\right)=-e^{-2t}u^2+h(t)e^{-t}u+c$

$e^{-2t}u\dfrac{du}{dt}-e^{-2t}u^2=-e^{-2t}u^2+h(t)e^{-t}u+c$

$e^{-2t}u\dfrac{du}{dt}=h(t)e^{-t}u+c$

$u\dfrac{du}{dt}=h(t)e^tu+ce^{2t}$