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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

Given $y(0) =2000$ and $\large\frac{dy}{dx} =3200-20y^2$ then find the value of $\lim _{x \to \infty} y(n)$

$(a)\;does\;not\;exist\\ (b)\;Given\;differential\;equation\; cannot\;have \;y(0)=zero \\ (c)\;-40 \\ (d)\;40 $

1 Answer

$\large\frac{dy}{1600-y^2}$$=20\; dx$
$\large\frac{dy}{y^2-1600}$$=-20 dx$
$\int \large\frac{(y+40)-(y-40)}{(y+40)(y-40)}$$dy=\int 20 \times 80 \;dx$
$\log (y- 40) -\log (y+40) =-1600 x+c$
$\log \bigg(\large\frac{y-40}{y+40}\bigg) $$=-1600 x +c$
$\log \bigg(\large\frac{y+40}{y-40}\bigg) $$=1600 x -c$
$y(0) =2000$
$\log \bigg( \large\frac{2040}{1960} \bigg) $$=1600 \times 0-c$
$-c= \log \bigg( \large\frac{51}{49}\bigg)$
$\log \bigg(\large\frac{y+40}{y-40}\bigg) $$=1600 x +\log \bigg(\large\frac{51}{49}\bigg)$
$\log \bigg(\large\frac{49 (y+40)}{51 (y-40)}\bigg)$$=1600x$
for $x +\infty $ L.HS should also $\to \infty$
So, $y-40 \to 0$
$y \to 40$
Hence d is the correct answer.
answered Feb 3, 2014 by meena.p
 

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