\(((3y^2-t^2)/y^5)dy/dx+t/(2y^4)=0\)

\(((3y^2-t^2)/y^5)dy/dx= -t/2y^4\)

\((3y^2-t^2)dy/dx= -(ty)/2\)

\((2t^2-6y^2)= ty dt/dy\)

\(dt/dy= (2t)/y-(6y)/t\) (1)

\(Put\ t/y= u => dt/dy= u+y (du)/dy\)

Hence the equation (1) will be

\(u+y (du)/dy= 2u-6/u\)

\(y (du)/dy= u-6/u\)

\(y (du)/dy= (u^2-6)/u\)

\((udu)/((u^2-6))= dy/y\)

Integrating on the both sides

\(1/2 \log(u^2-6)= \log(y)+\log(C)\)

\(u^2-6= Cy^2\)

\(t^2/y^2= Cy^2+6\)

\(t^2= y^2(Cy^2+6)\)

Use y(1)=1

1= 1(C+6)

C= 1-6

C= -5

\(t^2= y^2(6-5y^2)\)