Question

Find the particular solution of the differential equation `x (dy)/(dx) - y = x^2.e^x`, given y(1) = 0.

Answer 1

Given differential equation is `x (dy)/(dx) - y = x^2.e^x`

⇒ `(dy)/(dx) - y/x` = xe^x, which is of the form

`(dy)/(dx) + Py` = Q

Here, P = `-1/x` and Q = xe^x

I.F. = `e^(intpdx)`

= `e^(int (-1)/x dx)`

= `e^(-logx)`

= `e^(log  1/x)`

The solution is given by y.IF. = `intQ xx I.F.dx + C`

`y. 1/x = intxe^x xx 1/x dx + C`

`y/x = inte^xdx + C`

`y/x = e^x + C`

`y/x = e^x`  ...(i)

Given y = 0 when x = 1

From equation (i), we get

0 = 1.e¹ + C.1

⇒ C = –e

y = xe^x – ex