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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 $ln|\sqrt{1+u^2}|=ln|x|+c$

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  • $\begingroup$ As a start, switch to $y(-t)$ so that the LHS becomes $-(t\cdot y(-t))’$ $\endgroup$ – b00n heT Oct 4 '18 at 15:19
  • $\begingroup$ Since $u$ is a function of $t$ aswell you have to change $dy/dt$ to some kind of $du/dt$ which can be obtained by applying the rule for the derivative of a quotient. Furhter what exactly is $x$? $\endgroup$ – mrtaurho Oct 4 '18 at 20:23
  • $\begingroup$ Hint: $e^{ln x} = x$, and $e^C => C$. $\endgroup$ – Fund Monica's Lawsuit Oct 4 '18 at 20:53
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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.

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  • $\begingroup$ Could you check what we've done above please! $\endgroup$ – Taljana D Oct 4 '18 at 15:50
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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.

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  • $\begingroup$ Can you check what we've done above please. $\endgroup$ – Taljana D Oct 4 '18 at 15:51
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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?

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  • $\begingroup$ Can you check what we've done above please! $\endgroup$ – Taljana D Oct 4 '18 at 15:51
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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 $y(t)=t\cdot\sinh(\log t+c)$.

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  • $\begingroup$ This can be further simplified to $y=y_0-\frac{t^2}{4y_0}$ with $y_0\lt0$ $\endgroup$ – xidgel Oct 4 '18 at 20:18

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