# What is the general solution of the differential equation #y'-1/xy=2x^2\lnx#?

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(because Symbolab is addicted to weird #C# constants...)

(because Symbolab is addicted to weird

##### 1 Answer

# y = x^3lnx - x^3/2 + Cx #

#### Explanation:

We have:

# y'-1/x y=2x^2 lnx #

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) #

As the equation is already in this form, then the integrating factor is given by;

# I = e^(int P(x) dx) #

# \ \ = exp(int \ -1/x \ dx) #

# \ \ = exp(-lnx) #

# \ \ = exp(ln(1/x)) #

# \ \ = 1/x #

And if we multiply the DE by this Integrating Factor,

# (1/x) dy/dx - (1/x) 1/x y = (1/x) 2x^2 lnx #

# :. (1/x) dy/dx - (1/x^2) y = 2x lnx #

# :. d/dx (y/x) = 2x lnx #

This is now separable, so by *"separating the variables"* we get:

# y/x = 2 \ int \ x lnx \ dx #

To integrate the RHS integral we can apply integration by Parts:

Let

# { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=x, => v,=x^2/2 ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

We have:

# int \ (lnx)(x) \ dx = (lnx)(x^2/2) - int \ (x^2/2)(1/x) \ dx #

# :. int \ xlnx \ dx = (x^2lnx)/2 - int \ x/2 \ dx #

# " " = (x^2lnx)/2 - x^2/4 #

With this result, we have:

# y/x = 2 {(x^2lnx)/2 - x^2/4} + C #

Leading to the GS:

# y = 2x {(x^2lnx)/2 - x^2/4} + Cx #

# \ \ = x^3lnx - x^3/2 + Cx #