Find $x(t) $Given $x'(t), x(0)$

Question

"Solve $x'(t) = 2-tx^2 - t + 2x^2,$ with $x(0) = 0$"

Alright, so I assume that by "solve $x'(t)$" it means "find $x(t).$" To get $x(t)$ from $x'(t),$ I'll need the integral:

$\int 2 - tx^2 - t + 2x^2 dt$ = $2t - \frac{t^2 x^2}{2} - \frac{t^2}{2} + 2tx^2 + C$

From this point on, though, I'm a bit lost. If I just plug a $0$ in for every $t$ everything goes to zero anyways except for $C$, so I guess that could imply that $C$ is just $0$? That seems overly simplistic and like I'm skipping some steps, though. Would anyone be able to point me in the right direction on this?

Answer 1

You face a differential equation $$x' = 2-tx^2 - t + 2x^2$$ that is to say $$x'(t)=(2-t)(1+x^2)$$ which is separable and can write $$\frac {dx}{1+x^2}=(2-t)\,dt$$ Integrate both sides to get $$\tan^{-1}(x)=2t-\frac 12 t^2+C$$ from which $$x=\tan\left(2t-\frac 12 t^2+C \right)$$ Now, apply the condition to get the value of $C$.

Answer 2

To solve $x'(t) = 2-tx^2 - t + 2x^2,$ you must find a function $x$ such that it solves the equation, in this case $x(t)=\tan \left(\frac{1}{2}\left(c_1-t^2+4t\right)\right)$ is the solution and to find the $c_1$ you have to solve with the initial conditions given, in this case you want that:

$x(0)=\tan \left(\frac{1}{2}\left(c_1-(0)^2+4(0)\right)\right)=\tan \left(\frac{c_1}{2}\right)=0$ this equivalent to $tan^{-1}(0)=0=\frac{c_1}{2}$ which implies $c_1=0$ and the final solution is $x(0)=\tan \left(\frac{1}{2}\left(-t^2+4t\right)\right)$