# Differential equation with first derivative squared

How does one solve this type of differential equation?

$C_1(\dot{x})^2 + \dfrac{C_2}{C_3-x} = C_4$

Where the C's are all constants? Thanks.

• What if you simply and take square roots? $x'' = \pm \sqrt{...}$. – Moo Mar 6 '16 at 14:29
• Where is this monster coming from ? – Claude Leibovici Mar 6 '16 at 14:33
• It's a physics situation using kinetic energy and electric potential energy between two particles. Say we have 2 positively charged particles. One to the left which is moving to the right with some velocity. The particle to the right is motionless held fixed. the differential equation is for the motion of the left particle using conservation of energy. The left particle starts at x=0 at t=0 with some initial velocity. moving towards the second particle which is located at $x=C_3$ – Ameet Sharma Mar 6 '16 at 14:41
• I should be very surprised by an analytical solution. Consider the problem numerically. – Claude Leibovici Mar 6 '16 at 14:47
• Thus... $$\sqrt{\frac{a-x}{b-x}}\cdot\dot x=\pm c,$$ for some suitable constants $(a,b,c)$, that is, a separable equation which can be integrated to yield an implicit formula for $x(t)$ of the form $$\int_{x(0)}^{x(t)}\sqrt{\frac{a-z}{b-z}}\,dz=\pm ct.$$ – Did Mar 6 '16 at 17:09