Question

For the differential equation, find the general solution:

`x dy/dx +  2y= x^2 log x`

Answer 1

The given equation

`x dy/dx + 2y = x^2 log x`

or `dy/dx + (2/x)y = x log x`

Comparing with `dy/dx + Py = Q`,

P = `2/x` and Q = x log x

∴ I.F. = `e^(int P dx) = e^(int_x^2 dx)`

`= e^(2 log x) = e^(log x^2) = x^2`

Hence the required solution

∴ y × I.F. = ∫ Q × I.F. dx + C

⇒ y × x² = ∫ x² + x log x dx + C

⇒ x² y = ∫ x³ log x + C

⇒ x² y = `log x * x^4/4 - int 1/4 * x^4/4 dx + C`

⇒ x² y = `x^4/4 log x - 1/4 int x^3 dx + C`

⇒ x² y = `x^4/4 log x - 1/4 xx x^4/4 + C`

⇒ y = `x^2/16 (4 log x - 1) + C/x^2`

Which is the required solution.