If x = x(t) is the solution of the differential equation , x(0) = 2, then x(1) equals __________. [2024]

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Solution

Q.
If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

Ans.
(14)

We have, $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$

$\Rightarrow \frac{d x}{d t} = \frac{2 x}{t + 1} + {(t + 1)}^{3}; P = \frac{- 2}{t + 1}, Q = {(t + 1)}^{3}$

So, $I . F . = e^{\int P d t} = e^{\int \frac{- 2}{t + 1} d t} = e^{- 2 \log (t + 1)} = {(t + 1)}^{- 2}$

The Solution of given differential equation is given by $x (I . F .) = \int Q \cdot (I . F .) d t + C$

$\Rightarrow \frac{x}{{(t + 1)}^{2}} = \int \frac{{(t + 1)}^{3}}{{(t + 1)}^{2}} d t + C$

$\Rightarrow \frac{x}{{(t + 1)}^{2}} = \int (t + 1) d t + C \Rightarrow x = {(t + 1)}^{2} [\frac{t^{2}}{2} + t + C]$

Putting x(0) = 2, we get

$2 = {(1)}^{2} (0 + 0 + C) \Rightarrow C = 2$

$∴ x = {(t + 1)}^{2} (\frac{t^{2}}{2} + t + 2)$

At $t = 1, x = {(2)}^{2} (\frac{1}{2} + 1 + 2) = 4 (\frac{7}{2}) = 14$ .

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