solve-x-2-3-y-x-3-1-y-x-2-

Question Number 50430 by Abdo msup. last updated on 16/Dec/18

s o l v e (x^{2} + 3) y^{^{'}} + (x^{3} - 1) y = x^{2}

Commented by Abdo msup. last updated on 17/Dec/18

(h e) (x^{2} + 3) y^{^{'}} + (x^{3} - 1) y = 0 \Rightarrow

(x^{2} + 3) y^{^{'}} = - (x^{3} - 1) y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{x^{3} - 1}{x^{2} + 3} \Rightarrow

l n ∣ y ∣ = \int \frac{1 - x^{3}}{x^{2} + 3} d x + c = \int \frac{1 - x (x^{2} + 3 - 3)}{x^{2} + 3} d x + c

= \int \frac{1 + 3 x}{x^{2} + 3} d x - \int x d x + c

= \frac{3}{2} \int \frac{2 x}{x^{2} + 3} d x + \int \frac{d x}{x^{2} + 3} - \frac{x^{2}}{2} + c

= \frac{3}{2} l n (x^{2} + 3) - \frac{x^{2}}{2} + \int \frac{d x}{x^{2} + 3} b u t

\int \frac{d x}{x^{2} + 3} =_{x = \sqrt{3} u} \int \frac{\sqrt{3} d u}{3 (1 + u^{2})}

= \frac{1}{\sqrt{3}} a r c t a n (\frac{x}{\sqrt{3}}) \Rightarrow

l n ∣ y ∣ = \frac{3}{2} l n (x^{2} + 3) - \frac{x^{2}}{2} + \frac{1}{\sqrt{3}} a r c t a n (\frac{x}{\sqrt{3}}) + c \Rightarrow

y = K e^{- \frac{x^{2}}{2}} {(x^{2} + 3)}^{\frac{3}{2}} . e^{\frac{1}{\sqrt{3}} a r c t a n (\frac{x}{\sqrt{3}})} \dots b e c o n t i n u e d \dots