If $\mathrm{x}=\mathrm{x}(\mathrm{t})$ is the solution of the differential equation $(\mathrm{t}+1) \mathrm{dx}=\left(2 \mathrm{x}+(\mathrm{t}+1)^{4}\right) \mathrm{dt}, \mathrm{x}(0)=2$, then, $\mathrm{x}(1)$ equals ____.

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Differential Equations

x = x (t)

is the solution of the differential equation

(t + 1) dx = (2 x + (t + 1)^{4}) dt, x (0) = 2

, then,

x (1)

equals ____.

Answer: 14

(t + 1) d x = (2 x + (t + 1)^{4}) d t ⟹ \frac{d x}{d t} - \frac{2}{t + 1} x = (t + 1)^{3}

, a linear ODE. Integrating factor:

I F = e^{- \int \frac{2}{t + 1} d t} = \frac{1}{( t + 1 ) ^{2}}

\frac{x}{( t + 1 ) ^{2}} = \int (t + 1) d t = \frac{( t + 1 ) ^{2}}{2} + C ⟹ x = \frac{( t + 1 ) ^{4}}{2} + C (t + 1)^{2}

t = 0, x = 2

2 = \frac{1}{2} + C ⟹ C = \frac{3}{2}

.
At

t = 1

x (1) = \frac{2 ^{4}}{2} + \frac{3}{2} (2)^{2} = 8 + 6 = 14

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer

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