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,

\[\frac{dy}{dx} = e^x \sin x\]

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

Integrating both sides, we get

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

\[ \Rightarrow y = I + C . . . . . . . . . . \left( 1 \right)\]

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

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

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

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

\[ \Rightarrow I = \sin x e^x - \cos x e^x - I ...........\left[\text{From (2)} \right]\]

\[ \Rightarrow 2I = \sin x e^x - \cos x e^x \]

\[ \Rightarrow I = \frac{1}{2} e^x \left( \sin x - \cos x \right) . . . . . . . . . \left( 3 \right)\]

From (1) and (3) we get

\[ \therefore y = \frac{1}{2} e^x \left( \sin x - \cos x \right) + C . . . . . . . . . \left( 4 \right)\]

Now equation of the curve passes through (0, 0)

Therefore when x = 0; y = 0

Putting x = 0 and y = 0 in (4) we get

\[ \therefore 0 = \frac{1}{2} e^0 \left( \sin 0 - \cos 0 \right) + C\]

\[ \Rightarrow C = \frac{1}{2}\]

Substituting the value of `C` in (4), we get

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

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