Solve \( \frac{d y}{d x}+y x=y^{2} e^{x^{2} / 2} \sin x \)

We start with the differential equation \[ \frac{dy}{dx} + xy = y^2 e^{\frac{x^2}{2}} \sin x. \] This equation is a Bernoulli equation of the form \[ \frac{dy}{dx} + P(x)y = Q(x)y^n, \] with \(P(x)=x\), \(Q(x)=e^{\frac{x^2}{2}} \sin x\) and \(n=2\). ### Step 1. Bernoulli Substitution For a Bernoulli equation with \(n \neq 0,1\) we make the substitution \[ u=y^{1-n}=y^{1-2}=y^{-1}. \] Differentiate \(u\) with respect to \(x\): \[ \frac{du}{dx} = \frac{d}{dx}\left(y^{-1}\right) = -y^{-2}\frac{dy}{dx}. \] Solving for \(\frac{dy}{dx}\), we have \[ \frac{dy}{dx} = -y^2 \frac{du}{dx}. \] ### Step 2. Substitute into the Original Equation Substitute \(\frac{dy}{dx} = -y^2\frac{du}{dx}\) into the original equation: \[ -y^2 \frac{du}{dx} + xy = y^2 e^{\frac{x^2}{2}} \sin x. \] Divide the entire equation by \(y^2\) (assuming \(y\neq 0\)): \[ -\frac{du}{dx} + \frac{x}{y} = e^{\frac{x^2}{2}} \sin x. \] Since \(u=\frac{1}{y}\), then \(\frac{x}{y} = xu\), and the equation becomes \[ -\frac{du}{dx} + xu = e^{\frac{x^2}{2}} \sin x. \] Multiply the equation by \(-1\) to obtain a standard linear form: \[ \frac{du}{dx} - xu = -e^{\frac{x^2}{2}} \sin x. \] ### Step 3. Solve the Linear Equation for \(u\) The linear equation in \(u\) is \[ \frac{du}{dx} - xu = -e^{\frac{x^2}{2}} \sin x. \] The integrating factor \(\mu(x)\) is given by \[ \mu(x)=e^{\int -x\,dx}=e^{-\frac{x^2}{2}}. \] Multiply the entire equation by the integrating factor: \[ e^{-\frac{x^2}{2}}\frac{du}{dx} - xe^{-\frac{x^2}{2}} u = -e^{-\frac{x^2}{2}} e^{\frac{x^2}{2}} \sin x = -\sin x. \] Notice that the left-hand side is the derivative of \(e^{-\frac{x^2}{2}} u\): \[ \frac{d}{dx}\left(e^{-\frac{x^2}{2}} u\right) = -\sin x. \] Integrate both sides with respect to \(x\): \[ e^{-\frac{x^2}{2}} u = \int -\sin x\,dx. \] The integral on the right-hand side is \[ \int -\sin x\,dx = \cos x + C, \] where \(C\) is the constant of integration. Thus, \[ e^{-\frac{x^2}{2}} u = \cos x + C. \] Solve for \(u\): \[ u = e^{\frac{x^2}{2}} (\cos x + C). \] Recall that \(u=\frac{1}{y}\); hence, \[ \frac{1}{y} = e^{\frac{x^2}{2}} (\cos x + C). \] ### Step 4. Solve for \(y\) Taking the reciprocal, we obtain \[ y = \frac{1}{e^{\frac{x^2}{2}} (\cos x + C)} = \frac{e^{-\frac{x^2}{2}}}{\cos x + C}. \] Thus, the general solution is \[ y(x)=\frac{e^{-\frac{x^2}{2}}}{\cos x + C}. \]