Solve $x'(t)=(x(t))^2-t^2+1 $ How can we solve $$x'(t)=(x(t))^2-t^2+1 $$?I have tried to check whether it is Exact, separable, homogeneous, Bernoulli or not. It doesn't resemble to none of them. Who can help me. Thank you. The source of question is CEU entrance examination.

The non-linear DE $$x'=P(t)+Q(t)x+R(t)x^2$$ is called Ricatti's equation. If $x_1$ is a known particular solution of it, then we can have a family of solutions of the OE of the form $x(t)=x_1+u$ where $u$ is a solution of $$u'=Ru^2+(Q+2x_1R)u$$ or the linear ODE: $$w'+(Q+2x_1R)w=-R,~~~w=u^{-1}$$ Here, as @Ishan noted correctly, one particular solution of your OE is $x_1=t$ and $$R=1,~~Q=0,~~P=1-t^2$$ and so we first solve $$w'+(0+2t\times1)w=-1,~~w=u^{-1}$$ or $w'+2tw=-1$. The solution of latter OE is $$w=\frac{e^{t^2}}{C-\int_{t_0}^te^{k^2}dk}$$ and sofar we get $$x(t)=t+\frac{C-\int_{t_0}^te^{k^2}dk}{e^{t^2}}$$
