Question

Find the equation of the curve which passes through the origin and has the slope x + 3y − 1 at any point (x, y) on it.

Answer 1

According to the question,
$\frac{dy}{dx}=x+3y-1$

$\Rightarrow \frac{dy}{dx}-3y=x-1$
$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-3 \\ Q=x-1 \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{-\int 3dx}=e^{-3x} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{\mathrm{e}}^{-3\mathrm{x}}=\int \left(x-1\right)e^{-3x}dx+C \\ \Rightarrow y{e}^{-3x}=\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{e^{-3x}}dx-\int e^{-3x}dx+C \\ \Rightarrow y{e}^{-3x}=x\int e^{-3x}dx-\int \left[\frac{d}{dx}\left(x\right)\int e^{-3x}dx\right]dx-\int e^{-3x}dx+C \\ \Rightarrow y{e}^{-3x}=-\frac{1}{3}x{e}^{-3x}+\frac{1}{3}\int e^{-3x}dx-\int e^{-3x}dx+C \\ \Rightarrow y{e}^{-3x}=-\frac{1}{3}x{e}^{-3x}-\frac{1}{9}{e}^{-3x}+\frac{1}{3}{e}^{-3x}+C \\ \Rightarrow y=-\frac{1}{3}x-\frac{1}{9}+\frac{1}{3}+C{e}^{3x} \\ \Rightarrow y=-\frac{1}{3}x+\frac{2}{9}+C{e}^{3x} \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{throught}\mathrm{the}\mathrm{origin},\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}. \\ \Rightarrow 0=-0+\frac{2}{9}+C{e}^0 \\ \Rightarrow C=-\frac{2}{9} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve},\mathrm{we}\mathrm{get} \\ y=-\frac{1}{3}x+\frac{2}{9}\left(1-e^{3x}\right) \\ \Rightarrow y+\frac{1}{3}x=\frac{2}{9}\left(1-e^{3x}\right) \\ \Rightarrow 3\left(3y+x\right)=2\left(1-e^{3x}\right) \end{gathered}$$