Find the equation of a curve passing through the point

Solution:

The differential equation of the curve is:

$y^{\prime}=e^{x} \sin x$

$\Rightarrow \frac{d y}{d x}=e^{x} \sin x$

$\Rightarrow d y=e^{x} \sin x$

Integrating both sides, we get:

$\int d y=\int e^{x} \sin x d x$              ...(1)

$\int d y=\int e^{x} \sin x d x$

$\Rightarrow I=\sin x \int e^{x} d x-\int\left(\frac{d}{d x}(\sin x) \cdot \int e^{x} d x\right) d x$

$\Rightarrow I=\sin x \cdot e^{x}-\int \cos x \cdot e^{x} d x$

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

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

$\Rightarrow I=e^{x} \sin x-e^{x} \cos x-I$

$\Rightarrow 2 I=e^{x}(\sin x-\cos x)$

$\Rightarrow I=\frac{e^{x}(\sin x-\cos x)}{2}$

Substituting this value in equation (1), we get:

$y=\frac{e^{x}(\sin x-\cos x)}{2}+\mathrm{C}$            ...(2)

Now, the curve passes through point (0, 0).

$\therefore 0=\frac{e^{0}(\sin 0-\cos 0)}{2}+\mathrm{C}$

$\Rightarrow 0=\frac{1(0-1)}{2}+\mathrm{C}$

$\Rightarrow \mathrm{C}=\frac{1}{2}$

Substituting $\mathrm{C}=\frac{1}{2}$ in equation (2), we get:

$y=\frac{e^{x}(\sin x-\cos x)}{2}+\frac{1}{2}$

$\Rightarrow 2 y=e^{x}(\sin x-\cos x)+1$

$\Rightarrow 2 y-1=e^{x}(\sin x-\cos x)$

Hence, the required equation of the curve is $2 y-1=e^{x}(\sin x-\cos x)$.