Solve the differential equation #2xlnx dy/dx + y = 0#?
1 Answer
# y = A/sqrt(lnx) #
Explanation:
We have:
# 2xlnx dy/dx + y = 0# ..... [A]
We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;
# dy/dx + P(x)y=Q(x) #
So rewrite the equations in standard form as:
# dy/dx + 1/(2xlnx)y = 0 ..... [B] #
Then the integrating factor is given by;
# I = e^(int P(x) dx) #
# \ \ = exp(int \ 1/(2xlnx) \ dx) #
# \ \ = exp( 1/2ln|lnx| ) # (see notes at end)
# \ \ = exp( lnsqrt(|lnx|)) #
# \ \ = sqrt(lnx)) #
And if we multiply the DE [B] by this Integrating Factor,
# sqrt(lnx)dy/dx + sqrt(lnx)1/(2xlnx)y = 0 #
# :. d/dx (ysqrt(lnx)) = 0 #
# :. ysqrt(lnx)) = A #
Which we can rearrange to get:
# :. y = A/sqrt(lnx) #
Which, is the General Solution.