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

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

Toolbox:
• When the function appears in the form of integration of $udv$ then it can be integrated by parts which is $uv - \int vdu$
Step 1:
Given $\large\frac{dy}{dx}$$= e^x\sin x seperating the variables dy = e^x\sin x \int dy = \int e^x\sin x dx \int e^{x\sin x} dx can be done by the method of integration by parts. Let u = e^x; hence du = e^xdx and dv = \sin xdx; so v =- \cos x Step 2: uv - \int v du I = -(e^{x\cos x}) - \int (-\cos x).e^xdx \;\;= - (e^{x\cos x}) +\int e^{x\cos x}dx Again using the same method of integration Step 3: Let u = e^x ; du = e^xdx Let dv = \cos xdx; hence v = \sin x -(e^x\cos x) + e^{x\sin x} - \int e^{x\sin x} dx I = e^{x\sin x} - e^{\cos x} -I 2I = e^{\large x\sin x} - e^{\large\cos x} I =\large\frac{ [e^{\sin x }- e^{x\cos x}]}{2} y =\large\frac{ [e^{x\sin x} - e^{x\cos x}]}{2 }$$+ C$
Step 4:
Now to evaluate the value of C let us take $x=0$ and $y =0$