Solving $\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$

Question

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

Answer 1

With homogeneous equations, you have two choices for the substitution.

Here we have

$$(x^2+t^2)\,dx=xt\,dt $$

You made the substitution $x=yt$, but you also had the choice of letting $t=ux$. As a general rule, the substitution which makes the algebraic simplification easier also makes the integration easier. So lets investigate the substitution $t=ux$

\begin{eqnarray} (x^2+u^2x^2)\,dx&=&ux^2(udx+xdu)\\ x^2\,dx&=&ux^3\,du\\ \frac{1}{x}\,dx&=&u\,du\\ \ln|x|+c_1&=&\frac{u^2}{2}\\ \ln x^2+c_2&=&\frac{t^2}{x^2}\\ t^2&=&x^2(\ln x^2+c_2) \end{eqnarray}

The initial condition then gives $c_2=0$. So

$$ t^2=x^2\ln x^2 $$

Answer 2

You got that

$$\dfrac{1+y^2}{y^3}dy=-\frac{1}{t}dt$$

This implies that, putting $y'(t)=\frac{dy(t)}{dt}$,

$$\dfrac{1+y(t)^2}{y(t)^3}y'(t)=-\frac{1}{t}$$

where this equality is as functions of $t$. In particular, since they are the same function, both sides of the equality must have the same primitive, ie

$$\int \dfrac{1+y(t)^2}{y(t)^3}y'(t) dt=\int -\frac{1}{t} dt$$

(but not a definite integral!)

On the left side, you can distribute the denominator and get that $$\int \dfrac{1+y(t)^2}{y(t)^3}y'(t) dt=-\frac{1}{2}y^{-2}(t)+\ln(|y(t)|)+C$$

And the right side remains $$\int -\frac{1}{t} dt=-\ln(t)+C$$

Therefore the solution satisfies the implicit equation $$-\frac{1}{2}y^{-2}(t)+\ln(|y(t)|)=-\ln(t)+C$$

The problem is that the initial data is at $t=0$ where this equality is not defined. Note that the substitution $x=ty$ is not valid for the initial data you have because $1\neq x(0)=0\cdot y =0 $

Answer 3

$$\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$$ 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. $$x(0)=1 \implies C=0$$ $$\ln x^2= \dfrac{t^2}{x^2}$$