Solving differential equation with given values Given that $x =2$ at $t = 1$, how would I solve this differential equation?
$$\frac{dx}{dt}=2-\frac2{x^2} \tag 1$$
Usually I would separate the variables, but due to the subtraction I don't think I can do it this way.
 A: The given ordinary differential equation
$$\frac{dx}{dt}=2-\frac2{x^2}$$
can be rewritten as
$$\frac{1}{2-\frac2{x^2}}dx=dt$$
which is separable. To integrate the left-hand side, one can first apply long division to form
$$\frac{1}{2-\frac2{x^2}}=-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+\frac{1}{2}$$
A: Method of fractions and fractions decomposition works fine:
$$\frac{dx}{dt}=2-\frac2{x^2}$$
$$\frac{dx}{dt}=\frac {2x^2-2}{x^2}$$
$$\frac{dx}{2x^2-2}=\frac {dt}{x^2}$$
Substract fractions:
$$-\frac{dx-2dt}{2}=\frac {dx}{2x^2-2}$$
$$d(2t-x)=\frac {dx}{x^2-1}$$
$$2d(2t-x)= \left ( \frac 1 {x-1} -\frac 1 {x+1}\right )  {dx}  $$
Integrate
$$2(2t-x)+C= \ln \left | \frac  {x-1}  {x+1}\right |   $$
Initial condition gives 
$$x(1)=2 \implies C=-\ln 3$$
$$4t-2x-\ln 3= \ln \left | \frac  {x-1}  {x+1}\right |   $$
$$4t=2x+ \ln \left | \frac  {3(x-1)}  {x+1}\right |   $$
A: The equation is both separable and autonomous ($t$ does not appear explicitly).
$$\frac{2x^2}{x^2-1}dx=\left(2+\frac1{x-1}-\frac1{x+1}\right)=4dt$$
and
$$2(x-2)+\log\left|\frac{x-1}{x+1}\right|-\log\frac13=4(t-1).$$
