Question

Find one-parameter families of solution curves of the following differential equation:-

$$(x+y)\frac{dy}{dx}=1$$

Solve the following differential equation:-

$$(x+y)\frac{dy}{dx}=1$$

Answer 1

We have,
$$(x+y)\frac{dy}{dx}=1$$
$$\Rightarrow \frac{dy}{dx}=\frac{1}{x+y}$$
$$\Rightarrow \frac{dx}{dy}=x+y$$
$$\Rightarrow \frac{dx}{dy}-x=y.....\left(1\right)$$
Clearly, it is a linear differential equation of the form
$$\frac{dx}{dy}+Px=Q$$
where
$$P=-1$$
$$Q=y$$
$$\therefore I.F.=e^{\int Pdy}$$
$$=e^{\int -1dy}$$
$$=e^{-y}$$
$$\text{ Multiplying both sides of }(1)\text{ by }{e}^{-y},\text{ we get }$$
$$e^{-y}(\frac{dx}{dy}-x)=e^{-y}y$$
$$\Rightarrow e^{-y}\frac{dx}{dy}-e^{-y}x=e^{-y}y$$
Integrating both sides with respect to y, we get

$$\Rightarrow e^{-y}x=y\int e^{-y}dy-\int [\frac{d}{dy}\left(y\right)\int e^{-y}dy]dy+C$$
$$\Rightarrow e^{-y}x=-y{e}^{-y}-e^{-y}+C$$
$$\Rightarrow e^{-y}x+y{e}^{-y}+e^{-y}=C$$
$$\Rightarrow (x+y+1)e^{-y}=C$$
$$\Rightarrow (x+y+1)=C{e}^y$$
$$\text{ Hence, }(x+y+1)=C{e}^y\text{ is the required solution.}$$