Question

For  the following differential equation find the particular solution.

`dy/ dx = (4x + y + 1),

when  y = 1, x = 0

Answer 1

`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.