Question

Solve the following Linear differential equation:

`x ("d"y)/("d"x) + 2y - x^2 log x` = 0

Answer 1

The given differential equation may be written as

`x/x ("d"y)/("d"x) + (2y)/x = (xlogx)/x`

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

Where P = `2/x

Q = x log x

Thus, the given equation is linear.

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

= `"e"^(int 2/x  "d"x)`

e^logx = `"e"^(logx^2)`

= x²

So the required solution is

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

yx² = `int x log x x^2  "d"x + "c"`

yx² = `int x^3 logx  "d"x + "c"`

u = `log x int "dv" = int x^3  "d"x`

du = `1/x  "d"x`

v = `x^4/4`

`int"u"  "dv" = "uv" - int "v"  "du"`

yx² = `logx(x^4/4) - int x^4/4 * 1/x  "d"x + "c"`

= `logx (x^4/4) - 1/4 int x^3  "d"x + "c"`

= `log x(x^4/4) - 1/4(x^4/4) + "c"`

x²y = `(x^4/4) logx - x^4/16 + "c"`

Multiply by 16

16x²y = 4x⁴ log x – x⁴ + 16c is a required solution