Answer to Question #277396 in Differential Equations for dkwnsgsu

Question

(x^3 - x)dy/dx - (3x^2 - 1)y = x^5 - 2x^3 + x , y(1) = 1

Answer 1

$$y'-\dfrac{3x^2-1}{x^3-x}y=\dfrac{x^4}{x^2-1}-\dfrac{2x^2}{x^2-1}+\dfrac{1}{x^2-1}$$

Integrating factor

$$\mu(x)=e^{-\int\tfrac{3x^2-1}{x^3-x}dx}=\dfrac{1}{x^3-x}$$

$$\dfrac{1}{x^3-x}y'-\dfrac{3x^2-1}{(x^3-x)^2}y$$

$$=\dfrac{x^3}{(x^2-1)^2}-\dfrac{2x}{(x^2-1)^2}+\dfrac{1}{x(x^2-1)^2}$$

$$(\dfrac{1}{x^3-x}y)'=\dfrac{x^4-2x^2+1}{x(x^2-1)^2}$$

$$(\dfrac{1}{x^3-x}y)'=\dfrac{(x^2-1)^2}{x(x^2-1)^2}$$

$$(\dfrac{1}{x^3-x}y)'=\dfrac{1}{x}$$

Integrate

$$\dfrac{1}{x^3-x}y=\int \dfrac{1}{x}dx$$

$$\dfrac{1}{x^3-x}y=\ln (|x|)+c_1$$

$$y=(x^3-x)\ln (x)+c_1(x^3-x)$$

$y(1) = 1$

$$1=(1-1)(0)+c_1(1-1)$$

$$1=0, False$$

Therefore the given IVP does not have solution.