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

We have to find the equation of the curve that passes through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x\]

\[\therefore dy = e^x \sin x\ dx\]

Integarting both sides, we get

\[\int dy = \int e^x \sin x\ dx\]

\[ \Rightarrow y = \int e^x \sin x\ dx ...............(1)\]

\[ \Rightarrow y = e^x \int \sin x\ dx - \int\left\{ \frac{d}{dx}\left( e^x \right) \int\sin x\ dx \right\} dx\]

\[ \Rightarrow y = - e^x \cos x + \int e^x \cos x\ dx\]

\[ \Rightarrow y = - e^x \cos x + \left[ e^x \int \cos x\ dx - \int\left\{ \frac{d}{dx}\left( e^x \right) \int\cos x\ dx \right\}dx \right]\]

\[ \Rightarrow y = - e^x \cos x + e^x \sin x - \int e^x \sin x\ dx\]

\[ \Rightarrow y = - e^x \cos x + e^x \sin x - y + C..............\left[\text{Using (1)}\right]\]

\[ \Rightarrow 2y = e^x \left( \sin x - \cos x \right) + C ...............(2)\]

The curve passes through the point (0, 0)

When, `x = 0; y = 0`

Substituting the value of `x` and `y` in (2), we get

\[0 = 1\left( 0 - 1 \right) + C\]

\[ \Rightarrow C = 1\]

\[\text{ Substituting the value of C in }\left( 2 \right),\text{ we get }\]

\[2y = e^x \left( \sin x - \cos x \right) + 1\]

Required equation of curve is `2y = e^x (sin x - cos x) + 1`