Question

Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.

Answer 1

`dy/dx + xy = xy^2` 

∴ `dy/dx = xy(y - 1)`

∴ `dy/(y(y - 1)) = x  dx`

∴ `int  dy/(y(y - 1)) = int x  dx`

Let `1/(y(y - 1)) = A/y + B/(y - 1)`

∴ 1 = A(y – 1) + By

Put y = 1,

`\implies` B = 1,

Put y = 0,

`\implies` A = – 1

∴ `int((-1)/y + 1/(y - 1))dy = intx  dx`

∴  – log y + log(y – 1) = `x^2/2 + c`

∴ `log((y - 1)/y) = x^2/2 + c`  ...(1)

Put y = 4,

x = 1

∴ `log(3/4) = 1/2 + c`

∴ c = `log(3/4) - 1/2`

∴ From (1), the particular solution is

`log((y - 1)/y) = x^2/2 + log(3/4) - 1/2`

i.e. `log[(4(y - 1))/(3y)] = (x^2 - 1)/2`