The way this is set up is confusing me on how to start.
Solve the Bernoulli equation.
$$y' = \frac{t^2 +3y^2}{2ty}$$ $$t>0$$
Solve Bernoulli equation
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ordinary-differential-equations
differential
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0Take $T = \log(t)$ and $Y = y^2$ then $\frac{d Y}{d T} = e^{2T} + 3Y$ – 2017-02-12
1 Answers
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First let $u=y^2$. Then $u'=2yy'$, so $$ u' = 2y \frac{t^2+3y^2}{2ty} = t + \frac{3}{t}u. $$ This is now a first-order linear differential equation, so we look for an integrating factor, $ e^{\int -3/t \, dt} = e^{-3\log{t}} = t^{-3} $. Hence $$ (t^{-3}u)' = t^{-2}, $$ and integrating gives $$ u = -t^2 + At^3, $$ and substituting $y$ back in, $$ y^2 = At^3-t^2, $$ or $y = \pm t\sqrt{At-1}$, since $t>0$.