The value of $y a s t \to \infty$ for the following differential equation for an initial value of y(1) = 0 is
$(4 t^{2} + 1) \frac{d y}{d t} + 8 y t - t = 0$
0.125

Open in App

Solution

The correct option is A 0.125
Given $(4 t^{2} + 1) \frac{d y}{d t} + 8 y t - t = 0$

$\frac{d y}{d t} + \frac{8 t}{4 t^{2} + 1} y = \frac{t}{4 t^{2} + 1}$

$I F = e^{\int \frac{8 t}{4 t^{2} + 1} d t} = e^{l n (4 t^{2} + 1)} = 4 t^{2} + 1$

$S o l u t i o n i s, y (I . F) = \int I . F Q (t) d t + C$

Hence the general solution is,

$y (4 t^{2} + 1) = \frac{t^{2}}{2} + C$

Given, y(t =1) = 0, therefore,

$0 (4 \times 1^{2} + 1) = \frac{1^{2}}{2} + C$
$C = - \frac{1}{2}$

Hence the solution is,

$y (4 t^{2} + 1) = \frac{t^{2}}{2} - \frac{1}{2}$

$i . e ., y = \frac{t^{2} - 1}{2 (4 t^{2} + 1)} = \frac{1 - \frac{1}{t^{2}}}{8 + \frac{2}{t^{2}}}$
$y (\infty) = \frac{1}{8} = 0.125$

Suggest Corrections

Similar questions

y (t)

y (t) + . . y (t) = 0

. y (0) = 1

y (0) = 1,

t \geq 0

\frac{d x}{d t} = 3 x

x = 5

t = 0

x

t = 4

\frac{d^{2} y (t)}{d t^{2}} + 2 \frac{d y (t)}{d t} + y (t) = δ (t) w i t h y (t) |_{t = 0} = - 2 a n d \frac{d y}{d t} |_{t = 0} = 0.

\frac{d y}{d t} |_{t = 0}

\frac{d x}{d t} = 10 - 0.2 x

x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0.

Join BYJU'S Learning Program