If y = y(x) is the solution to the differential equation, $\frac {dy}{dx} + 2y tan x = sin x $, $y (\frac {\pi}{3})=0$, then the maximum value of the function y(x) over R is equal to:
(A) $\frac {1}{2} $
(B) $\frac {1}{8} $
(C) $-\frac {15}{4}$
(D) 8 Continue reading $\frac {dy}{dx} + 2y tan x = sin x $
$y (\frac {\pi}{3})=0$
${y_{\max }} = ?$
Let, $ \sqrt {x + y} = t$
So, $ \frac{1}{{2\sqrt {x + y} }}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$
Or, $\frac{1}{{2t}}\left( {1 + \frac{{dy}}{{dx}}} \right) = \frac{{dt}}{{dx}}$
$\therefore \frac{{dy}}{{dx}} = 2t\frac{{dt}}{{dx}} – 1$
From the original differential equation,
$2t\frac{{dt}}{{dx}} – 1 = t$
$\therefore 2t\frac{{dt}}{{dx}} = 1 + t$
Or, $\frac{{2t}}{{1 + t}}dt = dx$
$ \Rightarrow 2\int {\frac{{1 + t – 1}}{{1 + t}}dt} = \int {dx} $
$ \Rightarrow 2\int {\left( {1 – \frac{1}{{1 + t}}} \right)dt} = \int {dx} $
$ \Rightarrow 2t – 2\ln (1 + t) = x + C$
$ \Rightarrow 2\sqrt {x + y} – 2\ln (1 + \sqrt {x + y} ) = x + C$
Or, $\sqrt {x + y} – \ln (1 + \sqrt {x + y} ) = \frac{x}{2} + C’$
