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प्रश्न
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
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उत्तर
`dy/ dx = (4x + y + 1)` ..(i)
Put 4x + y + 1 = t …(ii)
Differentiating w.r.t. x, we get
`4 + dy/dx = dt/ dx`
∴ `dy/dx = dt/ dx - 4` .... (iii)
Substituting (ii) and (iii) in (i), we get
`dt/ dx – 4 = t`
∴ `dt/ dx = t + 4`
∴ `dt/ (t + 4) = dx`
Integrating on both sides, we get
`intdt/(t+4) = int dx`
∴ log | t + 4 | = x + c
∴ log |(4x + y + 1) + 4 | = x + c
∴ log | 4x + y + 5 | = x + c …(iv)
When y = 1, x = 0, we have
log |4(0) + 1 + 5| = 0 + c
∴ c = log |6|
Substituting c = log |6| in (iv), we get
log |4x + y + 5| = x + log |6|
∴ log |4x + y + 5 | - log |6| = x
∴ log `|(4x+y+ 5) /6 | = x`,
which is the required particular solution.
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