Solving an inseparable initial value problem Solve $t$$\frac{dx}{dt}$ = $x$ + $\sqrt{t^2 +x^2}$ with $x$(1) = 0.
I tried to use substitution of $t = ux$ and ended up getting down to ( $\frac{1}{u\sqrt{u^2+1}}$ + $\frac{1}{u}$ ) $du$ = -$\frac{1}{x}$ $dx$ but from here that means i have to integrate $\frac{1}{u\sqrt{u^2+1}}$ which isn't very pretty if I remember and then still I need to get back into terms of $x$ and $t$ so I can substitute in the initial values and get the final result.
So is there a simpler way to solve this or another substitution I can use?
 A: $$tx' = x+\sqrt{t^2 +x^2}$$
$$\frac {tx'-x}{t^2} = \frac 1 {t^2}\sqrt{t^2 +x^2}$$
$$ \left (\frac xt\right )' = \frac 1t\sqrt{1 +\frac {x^2}{t^2}}$$
It's separable.
$$ \int \frac {d\left (\frac xt\right )}{\sqrt{1 +\frac {x^2}{t^2}}} =\int  \frac {dt}t$$
Evaluate both integrals..
$$\text{arcsinh}(x/t)=\ln |t| +K$$
Since $K=0$
$$x(t)=t \sinh(\ln|t|)$$
A: Then you should get answer $$x=t \operatorname{sinh}\left( \log{(t)}\right)=\frac{{{t}^{2}}}{2}-\frac{1}{2}$$
A: Around $t=1$ the substitution $t=\sqrt{s}$ is a monotonous reparametrization. Set $x(t)=u(t^2)$, then $\dot x=2t\dot u(t^2)$ and
$$
t\dot x(t)-x(t)=2s\dot u(s)-u(s)=\sqrt{s+u(s)^2}\implies
4 s^2\dot u^2-4sx\dot x=s,
\\
u=s\dot u-\frac1{4\dot u}.
$$
This last formula is a Clairaut equation which has a solution family
$$
u(s)=Cs-\frac1{4C},~~x(t)=Ct^2-\frac1{4C}
$$
and an envelope satisfying $s+\frac1{4\dot u^2}=0$, which has no solution for $s=t^2\ge 0$.
The above solution family may contain spurious solutions for the original equation due to taking the square in the derivation.
Testing the solutions results in
$$
2Ct^2=Ct^2-\frac1{4C}+\sqrt{t^2+(Ct^2-\frac1{4C})^2}=Ct^2-\frac1{4C}+\left|Ct^2+\frac1{4C}\right|=\max\left(2Ct^2,-\frac1{4C}\right)
$$
which is true for $C>0$.
