Solving $\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$ I have started self-studying differential equations and I have come across the following initial value problem
$$\frac{dx}{dt}=\frac{xt}{x^2+t^2}, \quad x(0)=1$$
Now, since $f(t,x)=\frac{xt}{x^2+t^2}$ is such that $f(rt,rx)=f(t,x)$ for every $r\in\mathbb{R}\setminus\{0\}$, we can use the change of variables $y=\frac{x}{t}$ to rewrite it in the form
\begin{align}
y+t\frac{dy}{dt} &=\frac{t^2 y}{t^2(1+y^2)} \\
&=\frac{y}{1+y^2} \\
\implies t\frac{dy}{dt} &=\frac{y}{1+y^2}-y \\
&=-\frac{y^3}{1+y^2} \\
\implies \frac{dy}{dt} &= \left(-\frac{y^3}{1+y^2}\right)\cdot\frac{1}{t}
\end{align}
which is separable, and becomes:
\begin{align}
\left(\frac{1+y^2}{y^3}\right)dy &= -\frac{dt}{t} \\
\implies \int_{y_1}^{y_2} \left(\frac{1+y^2}{y^3}\right) dy &= -\int_{t_1}^{t_2} \frac{dt}{t} \\
\implies -\frac{1}{2y_2^2}+\ln \left\lvert \frac{y_2}{y_1} \right\rvert + \frac{1}{2y_1^2} &= -\ln \left\lvert \frac{t_2}{t_1} \right\rvert
\end{align}
but now I don't see how to go forward and find $y(t)$. Also, I integrated from a generic time $t_1$ to a generic time $t_2$ because the right hand side wouldn't have converged otherwise.
So, I would appreciate any hint about how to go forward in solving this IVP.
Thanks

$$\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$$
Multiply by $2x$:
$$2x{dx}=\frac{2x^2t}{x^2+t^2}dt$$
$${dx^2}=\frac{x^2}{x^2+t^2}dt^2$$
Substitute $u=x^2,v=t^2$:
$${du}=\frac{udv}{u+v}$$
This is a first order linear DE $uv'=u+v$ that you can also solv as this:
$$u{du}={udv}-vdu$$
$$\dfrac {du}u=\dfrac {udv-vdu}{u^2}$$
$$\dfrac {du}u=d \left (\dfrac       {v}{u}\right)$$
Integrate:
$$\ln u=\dfrac {v}{u}+C$$
$$\ln x^2= \dfrac{t^2}{x^2}+C$$
Apply initial condition.
$$x(0)=1  \implies C=0$$
$$\ln x^2= \dfrac{t^2}{x^2}$$
