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

Answers (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
\(t^2= y^2(6-5y^2)\)
0

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