general solution of the differential equation, \( d y / d x+x y=x \), is

To solve the differential equation: \[ \frac{dy}{dx} + x y = x \] we recognize it as a first-order linear ordinary differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] where \( P(x) = x \) and \( Q(x) = x \). ### Step 1: Determine the Integrating Factor The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int x \, dx} = e^{\frac{x^2}{2}} \] ### Step 2: Multiply Through by the Integrating Factor Multiply every term in the differential equation by \( \mu(x) \): \[ e^{\frac{x^2}{2}} \frac{dy}{dx} + x e^{\frac{x^2}{2}} y = x e^{\frac{x^2}{2}} \] This simplifies to: \[ \frac{d}{dx} \left( y e^{\frac{x^2}{2}} \right) = x e^{\frac{x^2}{2}} \] ### Step 3: Integrate Both Sides Integrate both sides with respect to \( x \): \[ y e^{\frac{x^2}{2}} = \int x e^{\frac{x^2}{2}} \, dx + C \] Let \( u = \frac{x^2}{2} \), hence \( du = x \, dx \). The integral becomes: \[ \int e^u \, du = e^u + C = e^{\frac{x^2}{2}} + C \] So, \[ y e^{\frac{x^2}{2}} = e^{\frac{x^2}{2}} + C \] ### Step 4: Solve for \( y \) Divide both sides by \( e^{\frac{x^2}{2}} \): \[ y = 1 + C e^{-\frac{x^2}{2}} \] ### Final Answer After simplifying, all solutions are 1 plus a constant times e^–½x². Thus, y = 1 + C e^−x²⁄2