Question

Differential equations

Answer 1

\[\sqrt{y^2+1}dx - xydy=0\]

Answer 2

I don't really know how to proceed to separate both y's and x's so I can be able to integrate them separately :[

Answer 3

\[\begin{align*}
\sqrt{y^2+1}\,dx-xy\,dy&=0\\\\
\sqrt{y^2+1}\,dx&=xy\,dy\\\\
\frac{dx}{x}&=\frac{y}{\sqrt{y^2+1}}\,dy
\end{align*}\]

Answer 4

ooo alright I think I know how to proceed after that, 1sec

Answer 5

\[\frac{ dx }{ x } - \frac{ y }{ \sqrt{y^2+1} }dy =0\] \[\int\limits_{}^{}\frac{ dx }{ x } - \int\limits_{}^{}\frac{ y }{ \sqrt{y^2 +1} }dy =C \] \[\ln|x| - \sqrt{y^2+1}=C\]

Answer 6

The \(y\) term is missing a small detail. When integrating, setting \(u=y^2+1\) would yield \(\dfrac{du}{2}=y\,dy\), not just \(du=y\,dy\).

Answer 7

oh true, I somehow did it twice and I canceled it haha

Answer 8

no wait, they cancel after you integrate

Answer 9

Ah yes you're right. One step ahead :P

Answer 10

yeah I pretty much simplified all and went to the last part. But is that the final answer? since this is separable type of ODE

Answer 11

Yes it's correct. You can always try your luck solving for \(y\) or \(x\) explicitly, but your final answer is good.

Answer 12

Alright, thanks. As for a Linear type of ODE. I will have to use the long formula where I have e^integral of A(x)dx .... blahablah right

Answer 13

Right. You could even write this ODE as a linear equation in \(x\) and see if you get the same answer.