Solving Riccati(?) Equation

Question

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

Answer 1

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 $$ \frac {1}{{(ve^{t^2})}}=\frac {\sqrt \pi}{2}\text {erf(t)}+K$$ $$ ve^{t^2}=\frac {1}{\frac {\sqrt \pi}{2}\text {erf(t)}+K}$$ $$\boxed{ \implies x(t)=\frac {e^{-t^2}}{\frac {\sqrt \pi}{2}\text {erf(t)}+K}+t}$$

Answer 2

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.

Answer 3

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.