# Solve differential equation ((3y^2-t^2)/y^5)dy/dx+t/(2y^4)=0, y(1)=1

Question
Solve differential equation $$((3y^2-t^2)/y^5)dy/dx+t/(2y^4)=0$$, y(1)=1

2021-02-06
$$((3y^2-t^2)/y^5)dy/dx+t/(2y^4)=0$$
$$((3y^2-t^2)/y^5)dy/dx= -t/2y^4$$
$$(3y^2-t^2)dy/dx= -(ty)/2$$
$$(2t^2-6y^2)= ty dt/dy$$
$$dt/dy= (2t)/y-(6y)/t$$ (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
$$t^2= y^2(6-5y^2)$$

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