Question

The particular solution of the differential equation `y(1 + log x) dx/dy - x log x = 0`, when x = e, y = e² is ______.

पर्याय

• y = ex log x

• ey = x log x

• xy = e log x

• y log x = ex

Answer 1

The particular solution of the differential equation `y(1 + log x) dx/dy - x log x = 0`, when x = e, y = e² is y = ex log x.

Explanation:

The given differential equation is 

 `y(1 + log x) [dx/dy] - x log x = 0`

`\implies ((1 + log x)dx)/(xlogx) = dy/y`

`\implies (1/(xlogx) + 1/x)dx = 1/y dy`

After integrating on both sides, we get

`int(1/(xlogx) + 1/x)dx = int 1/y dy`

Let log x = t

`\implies 1/x dx` = dt

So, `int1/t dt + int 1/x dx = int 1/y dy`

`\implies` log t + log x = log y + log c

`\implies` log tx = log yc

`\implies` tx = yc

`\implies` x log x = yc

When x = e then y = e²

Therefore, e log e = e²c

`\implies` e × 1 = e²c

`\implies` c = `1/e`

Hence, x log x = `y/e` `\implies` y = ex log x