What is a solution to the differential equation #xydy/dx-lnx=0# with y(1)=0?

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Answer 1 · 2017-01-17T17:23:00

# y^2 = ln^2x #

# xydy/dx-lnx = 0 #

Is a First Order Separable Differential Equation, so we can just separate the variables;

# \ \ \ xydy/dx = lnx #
# :. ydy/dx = lnx/x #

Leading to:

# int \ y \ dy =int \ lnx/x \ dx#

The LHS is immediately integrable, and for the RHS we use the substitution;

# u=ln x => (du)/dx = 1/x #

Which gives:

# int \ y \ dy =int \ u \ du #
# :. 1/2y^2 = 1/2u^2 + K #
# :. y^2 = u^2 + 2K #
# :. y^2 = ln^2x + 2K #

We are also given #y(1)=0 => #

# 0=ln1+2K #
# :. K=0 #

Hence the particular solution is;

# y^2 = ln^2x #

Check
We can quickly validate the solution;

# y^2 = ln^2x #
# 2ydy/dx = 2(lnx)*1/x #
# ydy/dx = (lnx)/x #
# xydy/dx = lnx #
# xydy/dx - lnx = 0 \ \ # QED