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 ((3y2t2)/y5)dy/dx+t/(2y4)=0, y(1)=1
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comentezq
Answered 2021-02-06 Author has 106 answers

((3y2t2)/y5)dy/dx+t/(2y4)=0
((3y2t2)/y5)dy/dx=t/2y4
(3y2t2)dy/dx=(ty)/2
(2t26y2)=tydt/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=2u6/u
y(du)/dy=u6/u
y(du)/dy=(u26)/u
(udu)/((u26))=dy/y
Integrating on the both sides
1/2log(u26)=log(y)+log(C)
u26=Cy2
t2/y2=Cy2+6
t2=y2(Cy2+6)
Use y(1)=1
1= 1(C+6)
C= 1-6
C= -5
t2=y2(65y2)

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