Question

Solve the following Linear differential equation:

`x ("d"y)/("d"x) + y = x log x`

Answer 1

The given differential equation may be written as

`x/x ("d"y)/("d"x) + y/x = (xlogx)/x`

`("d"y)/("d"x) + 1/x y = log x`

This is of the form `("d"y)/("d"x) + "P"y` = Q

Where P = `1/x

Q = log x

Thus, the given differential equation is linear.

I.F = `"e"^(int "Pd"x)`

= `"e"^(int 1/x  "d"x)`

= `"e"^(log x)`

= x

So, the solution of the given differential equation is given by

y × I.F = `int ("Q" xx "I.F")  "d"x + "c"`

yx = `int log xx x  "d"x + "c"`

yx = `int x log x  "d"x + "c"`

u = `log x int "dv" = int x  "d"x`

du = `1/x  "d"x`

v = `x^2/2`

`int "u"  "dv" = "uv" - int "v"  "du"`

yx = `logx(x^2/2) - int x^2/2 * 1/x  "d"x + "c"`

= `(x^2/2) log x - 1/2 int x  "d"x + "c"`

yx = `x^2/2 log x - 1/2(x^2/2) + "c"`

yx = `x^2/2 log x - x^2/4 + "c"`

Multiply by 4

4yx = 2x² log x – x² + 4c

4xy = 2x² log x – x² + 4c is a required solution.