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Solve the following $(x+y)^2 \large\frac{dy}{dx}$=$1$

$\begin{array}{1 1}(A) \;y + \tan (x+y) = c \\ (B)\; y + \tan^{-1} (x+y) = c\\(C) \;y - \tan (x+y) = c\\(D) \;y - \tan^{-1} (x+y) = c\end{array}$

1 Answer

  • 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:
$(x+y)^2 \large\frac{dy}{dx}$$=1$
Let $u= x+y$
$u^2\bigg[ \large\frac{du}{dx}$$-1\bigg]$$=1$
$\therefore \large\frac{du}{dx}$$-1=\large\frac{1}{u^2}$
$\qquad= \large\frac{1+u^2}{u^2}$
Step 2:
The variable have been separated
$\int \large\frac{u^2}{1+u^2} $$du =\int dx +c$
$I= \int \large\frac{u^2}{1+u^2} $$du$
$\quad= \int \large\frac{1+u^2-1}{1+u^2} $$du$
$\quad= \int \bigg( 1-\large\frac{1}{1+u^2}\bigg)$$du$
The GS is
$u- \tan ^{-1} u =x+c$
$(x+y)-\tan ^{-1}(x+y) =x+c$
$y-\tan^{-1} (x+y) =c$
answered Sep 4, 2013 by meena.p