How to find a one-family parameter of solutions of the differential equation xy' = y? Verify that each member of the family is a solution of the initial-value problem xy' = y, y(0) = 0. How do I approach these kind of problems?

Question

anonymous · Answer

By inspection* find a one-family parameter of solutions...

abb0t · Answer

First, get it in it's general form.

anonymous · Answer

Manipulating the expression we find $y'/y=1/x$ and we know $y'/y=(\log y)'$ ergo:$$(\log y)'=1/x\\log y=\int\frac1x\ dx=\log x+C\y=kx$$where $k=e^C$

anonymous · Answer

It is a separable equation.

anonymous · Answer

Alternatively you can just recall that $y'$ is effectively the slope of $y$; when we write $y=y'x$ it suggests that this function has constant slope :-)

anonymous · Answer

hmm:$$y'=y/x\\frac{dy}{dx}=\frac{y}x$$does this make it more obvious that the function must be a line thru the origin? :-)

anonymous · Answer

Yes it does. Thank you. :)