Question

Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\frac{dy}{dx}=e^x\sin x$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(0,0) \\ \mathrm{and}\mathrm{whose}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{dy}{dx}=e^x\sin x. \\ dy=e^x\sin xdx \\ \mathrm{Integarting}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int e^x\sin xdx \\ \Rightarrow y=\int e^x\sin xdx.....\left(1\right) \\ \Rightarrow y=e^x\int \sin xdx-\int \left\{\frac{d}{dx}\left(e^x\right)\int \sin xdx\right\}dx \\ \Rightarrow y=-e^x\cos x+\int e^x\cos xdx \\ \Rightarrow y=-e^x\cos x+\left[e^x\int \cos xdx-\int \left\{\frac{d}{dx}\left(e^x\right)\int \cos xdx\right\}dx\right] \\ \Rightarrow y=-e^x\cos x+e^x\sin x-\int e^x\sin xdx \\ \Rightarrow y=-e^x\cos x+e^x\sin x-y+C\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow 2y=e^x\left(\sin x-\cos x\right)+C.....\left(2\right) \\ \mathrm{The}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(0,0) \\ \mathrm{When},\mathrm{x}=0;\mathrm{y}=0 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathit{}\mathrm{and}y\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 0=1\left(0-1\right)+C \\ \Rightarrow C=1 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 2y=e^x\left(\sin x-\cos x\right)+1 \\ \mathrm{Required}\mathrm{equation}\mathrm{of}\mathrm{curve}\mathrm{is}2y=e^x\left(\sin x-\cos x\right)+1 \end{gathered}$$