Question

xdy/dx+ (x+1) y = x³

Answer 1

$x \dfrac{dy}{dx} + (x + 1)y = x^3 \\ \dfrac{dy}{dx} + (1 + \dfrac{1}{x})y = x^2 \, . . . (1) \\ \text{the integrating factor, I.F is}\\ e^{\int{(1 + \frac{1}{x}) dx}}\\ = e^{x + \ln x} \\ = xe^x\\ \text{multiply eq(1) through by} \, xe^x. \text{We have,} \\ xe^x \dfrac{dy}{dx} + xe^x(1 + \dfrac{1}{x})y = x^3 e^x\\ \dfrac{d}{dx}(xe^xy) = x^3 e^x\\ Integrating,\\ yx e^x = \int x^3 e^x + C . . . (2) \\ \text{By IBP, we have}\\ yx e^x = x^3 e^x - \int 3x^2 e^x dx \\ = x^3 e^x - 3 \int x^2 e^x \\ = x^3 e^x - 3 [x^2 e^x - \int 2x e^x dx] \\ = x^3 e^x - 3x^2 e^x + 6 \int x e^x dx \\ = x^3 e^x - 3x^2 e^x + 6[x e^x - \int e^x dx] \\ = x^3 e^x - 3x^2 e^x + 6xe^x - 6e^x \\ = e^x(x^3 - 3x^2 + 6x - 6) \\ \text{Thus, from eq(2). We have,} \\ yxe^x = e^x(x^3 - 3x^2 + 6x - 6) + C \\ y = \dfrac{e^x(x^3 - 3x^2 + 6x - 6)}{xe^x} + \dfrac{C}{x}e^{-x} \\ Hence,\\ y(x) = \dfrac{C}{x}e^{-x} + x^2 - 3x + 6 - \dfrac{6}{x}$