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(A) The given differential equation is $y^{\prime} = e^{x} \sin x$.This can be written as $\frac{dy}{dx} = e^{x} \sin x$.Integrating both sides with r

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

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(A) The given differential equation is $y^{\prime} = e^{x} \sin x$.This can be written as $\frac{dy}{dx} = e^{x} \sin x$.Integrating both sides with respect to $x$:$y = \int e^{x} \sin x \, dx + C$  ...........$(1)$To evaluate $I = \int e^{x} \sin x \, dx$,we use integration by parts:$I = \sin x \cdot e^{x} - \int \cos x \cdot e^{x} \, dx$$I = e^{x} \sin x - [\cos x \cdot e^{x} - \int(-\sin x) \cdot e^{x} \, dx]$$I = e^{x} \sin x - e^{x} \cos x - I$$2I = e^{x}(\sin x - \cos x)$$I = \frac{e^{x}(\sin x - \cos x)}{2}$Substituting this into equation $(1)$:$y = \frac{e^{x}(\sin x - \cos x)}{2} + C$  ...........$(2)$Since the curve passes through $(0,0)$,we substitute $x=0$ and $y=0$:$0 = \frac{e^{0}(\sin 0 - \cos 0)}{2} + C$$0 = \frac{1(0 - 1)}{2} + C$$0 = -\frac{1}{2} + C \Rightarrow C = \frac{1}{2}$Substituting $C = \frac{1}{2}$ into equation $(2)$:$y = \frac{e^{x}(\sin x - \cos x)}{2} + \frac{1}{2}$$2y = e^{x}(\sin x - \cos x) + 1$$2y - 1 = e^{x}(\sin x - \cos x)$.

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

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