For the differential equation, find the particular solution dydx = (4x +y + 1), when y = 1, x = 0 - Mathematics and Statistics

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Sum

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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Chapter 1.8: Differential Equation and Applications - Q.5

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