 # Solve differential equation ((3y^2-t^2)/y^5)dy/dx+t/(2y^4)=0, y(1)=1 ddaeeric 2021-02-05 Answered
Solve differential equation $\left(\left(3{y}^{2}-{t}^{2}\right)/{y}^{5}\right)dy/dx+t/\left(2{y}^{4}\right)=0$, y(1)=1
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$\left(\left(3{y}^{2}-{t}^{2}\right)/{y}^{5}\right)dy/dx+t/\left(2{y}^{4}\right)=0$
$\left(\left(3{y}^{2}-{t}^{2}\right)/{y}^{5}\right)dy/dx=-t/2{y}^{4}$
$\left(3{y}^{2}-{t}^{2}\right)dy/dx=-\left(ty\right)/2$
$\left(2{t}^{2}-6{y}^{2}\right)=tydt/dy$
$dt/dy=\left(2t\right)/y-\left(6y\right)/t$ (1)

Hence the equation (1) will be
$u+y\left(du\right)/dy=2u-6/u$
$y\left(du\right)/dy=u-6/u$
$y\left(du\right)/dy=\left({u}^{2}-6\right)/u$
$\left(udu\right)/\left(\left({u}^{2}-6\right)\right)=dy/y$
Integrating on the both sides
$1/2\mathrm{log}\left({u}^{2}-6\right)=\mathrm{log}\left(y\right)+\mathrm{log}\left(C\right)$
${u}^{2}-6=C{y}^{2}$
${t}^{2}/{y}^{2}=C{y}^{2}+6$
${t}^{2}={y}^{2}\left(C{y}^{2}+6\right)$
Use y(1)=1
1= 1(C+6)
C= 1-6
C= -5
${t}^{2}={y}^{2}\left(6-5{y}^{2}\right)$