Question

Find the particular solution of the differential equation `x^2 dy/dx + y^2 = xy dy/dx`, if y = 1 when x = 1.

Answer 1

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

∴ `(xy - x^2) dy/dx = y^2`

∴ `dy/dx = y^2/(xy - x^2)`

This is homogeneous differential equation.

Put y = `vx -> dy/dx`

= `v + x (dv)/dx`

Given equation becomes

`v + x (dv)/(dx) = (v^2x^2)/(x^2v - x^2)`

= `v^2/(v - 1)`

∴ `x (dv)/(dx) = v^2/(v - 1) - v `

= `(v^2 - v^2 + v)/(v - 1)`

∴ `x (dv)/dx = v/(v - 1)`

∴ `((v - 1))/v dv = dx/x`

Integrating both sides

`int ((v - 1)/v)dv = int dx/x`

∴ `int (1 - 1/v)dv = int dx/x`

∴ v – log v = log x + c

Put v = `y/x`

∴ `y/x - log(y/x)` = log x + c

∴ `y/x - log y + log x` = log x + c

∴ `y/x - logy` = c

Put x = 1,

y = 1,

∴ c = 1

∴ y – x log y = x