91Ó°ÊÓ

To confirm that the differential equation is exact, we need to check if the following condition holds for partial derivatives: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) Here, \(M(e^{y}+1) \cos x\) and \(N = e^{y} \sin x\). Let's find the partial derivatives: \(\frac{\partial M}{\partial y} = \frac{\partial((e^{y}+1) \cos x)}{\partial y} = (e^{y}) \cos x\) \(\frac{\partial N}{\partial x} = \frac{\partial(e^{y} \sin x)}{\partial x} = e^{y} \cos x\) As the condition holds (\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)), we have confirmed that the differential equation is exact.

Now, we need to find a potential function \(\Psi(x, y)\) such that \(\frac{\partial \Psi}{\partial x} = M\) and \(\frac{\partial \Psi}{\partial y} = N\). Let's first integrate M with respect to x: \(\int M dx = \int(e^{y}+1) \cos x dx\) Using u-substitution (let \(u(e^{y}+1)\), \(du = e^{y}dy\)), we have \(\int u \cos x dx \implies u\sin x+\int \sin x du\) Our expression is now: \((e^{y}+1) \sin x + F(y)\) Now, let's differentiate this expression with respect to y and compare it with N: \(\frac{\partial}{\partial y} [(e^{y}+1) \sin x + F(y)] = e^{y} \sin x + \frac{dF}{dy} = N\) This implies: \(\frac{dF}{dy} = 0\) Which means F(y) is a constant (let it be C). Therefore, \(\Psi(x, y) = (e^{y}+1) \sin x + C\)

Now that we have found the potential function (ignoring the constant), we can write the general solution of the exact differential equation as: \(\Psi(x, y) = (e^{y}+1) \sin x + C = k\), where k is the constant of integration. Thus, the general solution is: \((e^{y}+1) \sin x=k\)