Question

The solution of $\frac{dy}{dx}=\frac{{(x+y)}^2}{(x+y)(y-2)}$ is given by

• A. $(x+2{)}^4(1+\frac{2(y-2)}{x+2})=2k{e}^{\frac{2(y-2)}{x+2}}$
• B. $(x+2{)}^4(1+\frac{2(y-2)}{(x+2)}))=k{e}^{\frac{2(y-2)}{(x+2)}}$
• C. $(x+2{)}^3(1+\frac{2(y-2)}{(x+2)}))=k{e}^{\frac{2(y-2)}{(x+2)}}$
• D. $Noneofthese$

Answer 1

The correct option is B $(x+2{)}^4(1+\frac{2(y-2)}{(x+2)}))=k{e}^{\frac{2(y-2)}{(x+2)}}$

$\frac{dy}{dx}=\frac{(x+y{)}^2}{(x+2)(y-2)}$

$\frac{dy}{dx}=\frac{(x+2+y-2{)}^2}{(x+2)(y-2)}$

Put $X=x+2,Y=y-2$

$\frac{dY}{dX}=\frac{(X+Y{)}^2}{XY}$

Put $Y=VX\Rightarrow \frac{dY}{dX}=V+X\frac{dV}{dX}$

$\Rightarrow V+X\frac{dV}{dX}=\frac{(1+V{)}^2}{V}$

$\frac{V}{2V+1}dV=\frac{dX}{X}$

$\Rightarrow \int (1-\frac{1}{1+2V})dV=2\int \frac{dX}{X}$

$\Rightarrow V-\frac{1}{2}ln(1+2V)=2lnX+C$

$\Rightarrow X^4(1+\frac{2Y}{X})=C{e}^{\frac{2Y}{X}}$

$\Rightarrow (x+2{)}^4(1+\frac{2(y-2)}{x+2})=C{e}^{\frac{2(y-2)}{x+2}}$