Solving Riccati(?) Equation Solve an Ordinary Differential Equation:
$\frac{dx}{dt}=-x^2+1+t^2$
I suppose that it's a Riccati equation, but there is no given  $w(t)$, where $x(t)=w(t) + \frac{1}{u(t)}$ .
I've found out that $x_1=\frac{1}{C_1+t}$, and what should I do next? I'm not sure if it's the right way anyway. I should find the $w(t)$ but don't know how.
I will be gratefull for any help
 A: Making the substitution 
$$
x = \frac{y'}{y}
$$
the new DE reads
$$
\frac{y''(t)-(1+t^2) y(t)}{y(t)} = 0
$$
or considering $y(t) \ne 0$
$$
y''(t)-(1+t^2) y(t)=0
$$
which is a linear DE.
A: By inspection one can see that $w(t)=t$ is a solution. Applying the usual process $x(t)=w(t)+\frac1{u(t)}$ gives
$$
1-\frac{\dot u}{u^2}=-t^2-2\frac{t}{u}-\frac1{u^2}+1+t^2
$$
so that
$$
\dot u=2tu+1
$$
which is a simple linear ODE.
A: For your substitution you need to know a particular solution to the equation..  Here you can try $x(t)=t$ and use your substitution ...
Another approach
$$\frac{dx}{dt}=-x^2+1+t^2$$
$$x'-1=-(x^2-t^2)$$
$$(x-t)'=-(x-t)(x+t)$$
Substitute $v=x-t$
$$v'=-v(v+2t)$$
$$v'+2vt=-v^2$$
Thats a Bernouilli's equation
$$(ve^{t^2})'=-v^2e^{t^2}$$
$$(ve^{t^2})'=-v^2e^{2t^2}e^{-t^2}$$
Integrate
$$\int \frac {dve^{t^2}}{{(ve^{t^2})}^2}=-\int e^{-t^2}dt$$
You need the error function 
$$ \frac {1}{{(ve^{t^2})}}=\frac {\sqrt \pi}{2}\text  {erf(t)}+K$$
$$ ve^{t^2}=\frac {1}{\frac {\sqrt \pi}{2}\text  {erf(t)}+K}$$
$$\boxed{ \implies  x(t)=\frac {e^{-t^2}}{\frac {\sqrt \pi}{2}\text  {erf(t)}+K}+t}$$
