Solve differential equation(x^2lnx)y'+xy=0

snowlovelydayM 2020-11-26 Answered

Solve differential equation (x2lnx)y+xy=0

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Expert Answer

Bella
Answered 2020-11-27 Author has 81 answers

Use the variable separation method to solve the differential equation
1/xdx=ln(x)+c
(x2lnx)y+xy=0
(x2lnx)dy/dx=xy
dy/y=x/(x2lnx)dx
dy/y=1/(xlnx)dx
integrated both sides
ln(y)=ln(lnx)+ln(c)
ln(y)=ln(lnx)1+ln(c)=ln(1/(ln(x)))+ln(c) :lna+lnb=lnab
:(1/(xlnx)dx),(putlnx=t1/xdx=dt),(1/xdt=ln(t)+c),(put the value of t we get1/(xlnx)dx=ln(ln(x))+c):
y=c/(ln(x))

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Jeffrey Jordon
Answered 2021-12-25 Author has 2047 answers

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