How do I "guess" a one-parameter family of solutions to the differential equation y' = (2y)/t?

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- Apr 5th 2013, 03:16 PMdspinneyone-parameter family of solutions
How do I "guess" a one-parameter family of solutions to the differential equation y' = (2y)/t?

- Apr 5th 2013, 03:33 PMProve ItRe: one-parameter family of solutions
Why guess? Just solve the DE, it's separable.

- Apr 5th 2013, 03:44 PMdspinneyRe: one-parameter family of solutions
That's what I was thinking... this class is so new to me and the instructor isn't always so good with teaching us all that we need to know. So I solved it and got 2ln(t) + C... this would be the one-parameter family?

- Apr 11th 2013, 07:53 AMHallsofIvyRe: one-parameter family of solutions
Yes, and what is the parameter? In other words, which value do you change to get a different solution to that same differential equation?

- Apr 11th 2013, 06:11 PMProve ItRe: one-parameter family of solutions
No, that's incorrect. You have only integrated one side after separating. Also, $\displaystyle \displaystyle \int{\frac{1}{x}\,dx} = \ln{|x|} + C$. It should be

$\displaystyle \displaystyle \begin{align*} \frac{dy}{dt} &= \frac{2y}{t} \\ \frac{1}{y}\,\frac{dy}{dt} &= \frac{2}{t} \\ \int{\frac{1}{y}\,\frac{dy}{dt}\,dt} &= \int{\frac{2}{t}\,dt} \\ \int{\frac{1}{y}\,dy} &= 2\ln{|t|} + C_1 \\ \ln{|y|} +C_2 &= \ln{\left| t^2 \right|} + C_1 \\ \ln{|y|} - \ln{\left| t^2 \right|} &= C \textrm{ where } C = C_1 - C_2 \\ \ln{\left| \frac{y}{t^2} \right| } &= C \\ \left| \frac{y}{t^2} \right| &= e^C \\ \frac{y}{t^2} &= A \textrm{ where } A = \pm e^C \\ y &= A\,t^2 \end{align*}$