Solving $\frac{d \xi(t)}{dt} = 1 + (\xi(t))^2$

Question

How should I go about solving $\frac{d \xi(t)}{dt} = 1 + (\xi(t))^2$.

So far I have noticed it has the form of a Riccati equation, which tells us that a solution is of the form $\xi(t) = \frac{u'}{u}$ where $u$ solves $$ u'' - (q_1 + \frac{q_2'}{q_2})u' + q_2q_0 u = 0,$$ (using the notation of the wiki) which gives us $$u'' + xu =0.$$

But this causes me problems to solve.

(This question comes about whilst trying to solve the PDE $$(1+x^2)u_x + u_y =0, \quad u(0,y) = g(y) $$ by the method of characteristics, $\xi(t)$ is intended to be the characteristic line.)

Answer 1

It is a separable equation: $\frac{\mathrm d\xi}{\mathrm dt}=1+\xi^2$. So$$\frac{\mathrm d\xi}{1+\xi^2}=\mathrm dt$$and therefore$$\arctan\xi=t+C$$for some constant $C$. Can you take it from here?

Answer 2

$$(1+x^2)u_x+u_y=0$$ $$\implies z=c_1$$ And we have: $$\frac {dx}{x^2+1}=\frac {dy}{1}$$ $$\int \frac {dx}{x^2+1}=\int {dy}$$ $$\arctan (x)=y+c_2$$ $$c_1=f(c_2) \implies u(x,y)=f(\arctan (x)-y)$$ $$ u(0,y)=g(y) \implies u(x,y)=g(y-\arctan (x))$$

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$$u'' - (q_1 + \frac{q_2'}{q_2})u' + q_2q_0 u = 0,$$ Since $q_2q_0 =1$ and $q_1=0$ $$u'' + u = 0,$$ According to Wiki you have since $(R=0,S=1)$ $$u''+u=0$$