Solving the IVP given by $\dot x=\frac{t-x}{t+x}$ and $x(0)=1$

Question

Find all solutions for $\dot x=\frac{t-x}{t+x}$ with $x(0)=1$.

I am seriously struggling to separate the variables since the fraction is quite complex. How may I be able to separate $x$ and $t$?

Answer 1

Hint...the method is fairly standard: substitute $x=vt$

You get a separable variable differential equation in $v$ and $t$

Answer 2

the general method is to change variable $x = vt$ as suggested by quinn. but for this problem, you can try this: $$0=(t+x)\, dx - (t-x)\, dt = t \, dx + x \ dt + x \, dx - t \ dt = d\left(tx+\frac12 (x^2-t^2)\right) $$ that is $$tx+\frac12 (x^2-t^2) = constant=0+\frac12.$$

Answer 3

If you perform the substitution z=t+x, you get $\dot x=\dot z -1$ ,so, after a few computations, you can find that $\dot z= \frac{2t}{z}$, so the new equation is with separable variables and, since $z_0=0+1=1> 0$, you have $z^2=z_0^2+2t^2$,$z_0=1$ => $z=\sqrt{1+2t^2}$ => $x(t)=z(t)-t=\sqrt{1+2t^2}-t$ $\forall t\in \Bbb R$

($t+x=\sqrt{1+2t^2}>=1>0)$. EDIT: Applying the theorem of existence and uniqueness of the solution, which holds since t-x/t+x is $C^1$ in a neghbourhood of (0,1), you have that this is the unique maximal solution.