Question

The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Answer 1

According to the question,
$\frac{dy}{dx}=x+y$
$\Rightarrow \frac{dy}{dx}-y=x$
$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-1 \\ Q=x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{-\int dx}=e^{-x} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{e}^{-x}=\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{e^{-x}}dx+C \\ \Rightarrow y{e}^{-x}=x\int e^{-x}dx-\int \left[\frac{d}{dx}\left(x\right)\int e^{-x}dx\right]dx+C \\ \Rightarrow y{e}^{-x}=-x{e}^{-x}+\int e^{-x}dx+C \\ \Rightarrow y{e}^{-x}=-x{e}^{-x}-e^{-x}+C \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{throught}\mathrm{the}\mathrm{origin},\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}. \\ \Rightarrow 0e^0=-0e^0-e^0+C \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve},\mathrm{we}\mathrm{get} \\ y{e}^{-x}=-x{e}^{-x}-e^{-x}+1 \\ \Rightarrow y{e}^{-x}+x{e}^{-x}+e^{-x}=1 \\ \Rightarrow \left(y+x+1\right)e^{-x}=1 \\ \Rightarrow \left(x+y+1\right)=e^x \end{gathered}$$