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Question
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
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Solution
`("d"y)/("d"x)` = (4x + y + 1) ......(i)
Put 4x + y + 1 = t .....(ii)
Differentiating w.r.t. x, we get
`4 + ("d"y)/("d"x) = ("dt")/("d"x)`
∴ `("d"y)/("d"x) = "dt"/("d"x) - 4` ......(iii)
Substituting (ii) and (iii) in (i), we get
`"dt"/("d"x) - 4` = t
∴ `"dt"/("d"x)` = u + 4
∴ `"dt"/("t" + 4)` = dx
Integrating on both sides, we get
`int "dt"/("t" + 4) = int "d"x`
∴ log |t + 4| = x + c
∴ log |(4x + y + 1) + 4| = x + c
∴ log |4x + y + 5| = x + c ......(iv)
When y = 1, x = 0
∴ log |4(0) + 1 + 5| = x + c
∴ c = log |6|
∴ log |4x + y + 5| = x + log 6 ....[From (iv)]
∴ log |4x + y + 5| – log 6 = x
∴ `log|(4x + y + 5)/6|` = x, which is the required particular solution
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