# 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.