Find the general solution of #y^2/x dy/dx = ln x# ?

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Answer 1 · 2018-06-10T12:11:34

# :. 4y^3=3x^2(2lnx-1)+k, or, #

# 4y^3=3x^2*ln(x^2/e)+k.#

# y^2/xdy/dx=lnx#.

#:. y^2dy=xlnxdx..............."[separable variable]"#.

#:." Integrating, "inty^2dy=intxlnxdx+c#.

#:. y^3/3=intxlnxdx+c...................(ast)#.

To find #intxlnxdx#, we use the Rule of Integration by Parts (IBP) :

IBP : #intuv'dx=uv-intu'vdx#.

We take, #u=lnx, v'=x. :. u'=1/x, v=intv'dx=x^2/2#.

#:. intxlnxdx=(lnx)(x^2/2)-int{(1/x)(x^2/2)}dx#,

#=x^2/2lnx-1/2intxdx#,

#:. intxlnxdx=x^2/2lnx-x^2/4#.

Utilising this in #(ast)#, we get the general solution :

#y^3/3=x^2/2lnx-x^2/4+c#.

# :. 4y^3=3x^2(2lnx-1)+k, or, #

# 4y^3=3x^2*ln(x^2/e)+k, "where, "k=12c.#