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प्रश्न
Find the particular solution of the differential equation `x (dy)/(dx) - y = x^2.e^x`, given y(1) = 0.
उत्तर
Given differential equation is `x (dy)/(dx) - y = x^2.e^x`
⇒ `(dy)/(dx) - y/x` = xex, which is of the form
`(dy)/(dx) + Py` = Q
Here, P = `-1/x` and Q = xex
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 = xex – ex
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