# 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

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:
$(x+y)^2 \large\frac{dy}{dx}$$=1 Let u= x+y \large\frac{dy}{dx}$$=1+\large\frac{dy}{dx}$
$u^2\bigg[ \large\frac{du}{dx}$$-1\bigg]$$=1$
$\therefore \large\frac{du}{dx}$$-1=\large\frac{1}{u^2} \large\frac{du}{dx}=\frac{1}{u^2}$$+1$
$\qquad= \large\frac{1+u^2}{u^2}$
$\large\frac{u^2}{1+u^2}$$du=dx Step 2: The variable have been separated \int \large\frac{u^2}{1+u^2}$$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$

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