Question

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e^2x.

Answer 1

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \frac{dy}{dx}=y+2x \\ \Rightarrow \frac{dy}{dx}-y=2x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=-1\mathrm{and}Q=2x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int dx} \\ =e^{-x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^{-x},\mathrm{we}\mathrm{get} \\ {e}^{-x}\left(\frac{dy}{dx}-y\right)=e^{-x}2x \\ \Rightarrow e^{-x}\frac{dy}{dx}-e^{-x}y=e^{-x}2x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{e}^{-x}=2\int \underset{\mathrm{II}}{e^{-x}}\underset{\mathrm{I}}{x}dx+C \\ \Rightarrow y{e}^{-x}=2x\int e^{-x}dx-2\int \left[\frac{d}{dx}\left(x\right)\int e^{-x}dx\right]dx+C \\ \Rightarrow y{e}^{-x}=-2x{e}^{-x}-2e^{-x}+C.....\left(2\right) \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{origin},\mathrm{we}\mathrm{have} \\ 0\times e^0=-2\times 0\times e^0-2e^0+C \\ \Rightarrow C=2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y{e}^{-x}=-2x{e}^{-x}-2e^{-x}+2 \\ \Rightarrow y=-2x-2+2e^x \\ \Rightarrow y+2\left(x+1\right)=2e^x \\ \mathrm{DISCLAIMER}:\mathrm{In}\mathrm{the}\mathrm{question}\mathrm{it}\mathrm{should}\mathrm{be}{\mathrm{e}}^x\mathrm{instead}\mathrm{of}{\mathrm{e}}^{2x}. \end{gathered}$$