Question

Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer 1

According to the question,
$\frac{dy}{dx}+5=x+y$
$\Rightarrow \frac{dy}{dx}-y=x-5$
$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-1 \\ Q=x-5 \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{-\int dx}=e^{-x} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{e}^{-x}=\int \left(x-5\right)e^{-x}dx+C \\ \Rightarrow y{e}^{-x}=\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{e^{-x}}dx-5\int e^{-x}dx+C \\ \Rightarrow y{e}^{-x}=x\int e^{-x}dx-\int \left[\frac{d}{dx}\left(x\right)\int e^{-x}dx\right]dx+5e^{-x}+C \\ \Rightarrow y{e}^{-x}=-x{e}^{-x}+\int e^{-x}dx+5e^{-x}+C \\ \Rightarrow y{e}^{-x}=-x{e}^{-x}-e^{-x}+5e^{-x}+C \\ \Rightarrow y{e}^{-x}=-x{e}^{-x}+4e^{-x}+C \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{throught}\mathrm{the}\mathrm{point}\left(0,2\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}. \\ 2e^0=-0e^0+4e^0+C \\ \Rightarrow C=-2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve},\mathrm{we}\mathrm{get} \\ y{e}^{-x}=-x{e}^{-x}+4e^{-x}-2 \\ \Rightarrow y=-x+4-2e^x \\ \Rightarrow y=4-x-2e^x \end{gathered}$$