Modelling a horizontal mass damper using differential equations

Question

I am trying to solve this second-order differential equation:

$y'' + y'+ y + (y')^2 =0$

I was able to solve the equation $y'' + y'+ y $, by substituting $y$ as $Ae^{kt}$. But now I have this new term $(y')^2$.

Note: This equation represents the simplified equation of forces on a horizontal mass damper.

Answer 1

Let $u=\dfrac{dy}{dt}$ ,

Then $\dfrac{d^2y}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dy}\dfrac{dy}{dt}=u\dfrac{du}{dy}$

$\therefore u\dfrac{du}{dy}+u^2+u+y=0$

$(y+u^2+u)\dfrac{dy}{du}=-u$

This belongs to an Abel equation of the second kind.

Let $v=y+u^2+u$ ,

Then $y=v-u^2-u$

$\dfrac{dy}{du}=\dfrac{dv}{du}-2u-1$

$\therefore v\left(\dfrac{dv}{du}-2u-1\right)=-u$

$v\dfrac{dv}{du}-(2u+1)v=-u$

$v\dfrac{dv}{du}=(2u+1)v-u$

Let $r=u+\dfrac{1}{2}$ ,

Then $v\dfrac{dv}{dr}=2rv-r+\dfrac{1}{2}$

Let $s=r^2$ ,

Then $\dfrac{dv}{dr}=\dfrac{dv}{ds}\dfrac{ds}{dr}=2r\dfrac{dv}{ds}$

$\therefore2rv\dfrac{dv}{ds}=2rv-r+\dfrac{1}{2}$

$v\dfrac{dv}{ds}=v-\dfrac{1}{2}+\dfrac{1}{4r}$

$v\dfrac{dv}{ds}=v-\dfrac{1}{2}\pm\dfrac{1}{4\sqrt s}$

This belongs to an Abel equation of the second kind in the canonical form.