Question

Find the particular solution of:

1.) x^2 y' -2xy=x^4 +3; where y = 2 and x = 1

Answer 1

$$x^2y'-2xy=x^4+3$$

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

$$I.F.=e^{\int(-2/x)dx}=\dfrac{1}{x^2}$$

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

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

Integrate

$$\int d(\dfrac{y}{x^2})=\int(1+\dfrac{3}{x^4})dx$$

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

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

$y = 2,x = 1$

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

$$C=2$$

The particular solution is

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