Question

Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer 1

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