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Solve the following $\large\frac{dy}{dx}$$=\sin(x+y) $

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Toolbox:
  • First order , first degree DE
  • Variable separable : Variables of a DE are rearranged to separate then, ie
  • $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • Can be written as $ \large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
$\large\frac{dy}{dx}$$=\sin (x+y)$
Let $u=x+y$
$\large\frac{du}{dx}$$=1+\large\frac{dy}{dx}$
The DE becomes
$\large\frac{du}{dx}$$-1=\sin u$
or $\large\frac{du}{1+\sin u}$$=dx$
Step 2:
The variables have been separated
$\int \large\frac{dx}{1+\sin u}$$=\int dx+c$
$I= \int \large\frac{du}{1+\sin u}$
$\quad= \int \large\frac{(1-\sin u)}{(1-\sin u)(1+\sin u)}$$du$
$\quad=\int \large\frac{1-\sin u}{1-\sin ^2 u}$$du$
$\quad=\int \large\frac{1-\sin u}{\cos^2 u}$$du$
$\quad= \int (\sec^2 u -\tan u \sec u )du$
$\quad= \tan u -\sec u$
$\therefore$ the GS is
$\tan u - \sec u = x+c$
or $\tan (x+y)-\sec (x+y)=x+c$
answered Sep 4, 2013 by meena.p
 
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