Use an appropriate change of variables to solve the differential equation. Use an appropriate change of variables to solve the differential equation.
$$t\frac{dy}{dt}-y=\sqrt{t^2+y^2}$$
My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.
Any help would be appreciated especially if you could help us with step by step.
Thanks!
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UPDATE:
$$t\frac{dy}{dt}-y=\sqrt{t^2+y^2}$$
$$\frac{dy}{dt}-\frac{y}t=\sqrt{1+\frac{y^2}{t}}$$ where u= y/t
$$\frac{dy}{dt}=\sqrt{1+u^2}+u=f(u)$$
$$f(u)-u=\sqrt{1+u^2}+u-u=\sqrt{1+u^2}$$
...
We ended up with $ln|\sqrt{1+u^2}|=ln|x|+c$
 A: Divide by $t$ to obtain,
\begin{equation}
\frac{dy}{dt} - \frac{y}{t} = \sqrt{1+\left ( \frac{y}{t} \right )^2},
\end{equation}
then use a change of variables $u = \frac{y}{t}$. The transformed equation should be integrable using standard methods.
A: HINT
One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = \sqrt{1+z^2}$$ which is separable.
A: Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$t\left(t\frac{dv(t)}{dt}+v(t)\right)-tv(t)=\sqrt{t^2+t^2v(t)^2}$$
Can you proceed?
A: By using the substitution $\displaystyle u=\frac yt$ - and therefore $\displaystyle \frac{du}{dt}=\frac{ty'-y}{t^2}$ - the differential equation becomes
$$\frac{du}{dt}=\frac1t\sqrt{1+u^2}$$
which can be solved by separation of variables. To be exact 
$$\begin{align}
\frac{du}{dt}&=\frac1t\sqrt{1+u^2}\\
\frac{du}{\sqrt{1+u^2}}&=\frac{dt}t\\
\int\frac{du}{\sqrt{1+u^2}}&=\int\frac{dt}t\\
\operatorname{arsinh}(u)&=\log t +c\\
u&=\sinh(\log t+c)
\end{align}$$
and therefore by resubstitution $y(t)=t\cdot\sinh(\log t+c)$.
