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Math Help - one-parameter family of solutions

  1. #1
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    one-parameter family of solutions

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

    Why guess? Just solve the DE, it's separable.
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    Re: 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?
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    Re: 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?
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    Re: one-parameter family of solutions

    Quote Originally Posted by dspinney View Post
    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?
    No, that's incorrect. You have only integrated one side after separating. Also, \displaystyle \int{\frac{1}{x}\,dx} = \ln{|x|} + C. It should be

    \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*}
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