solve y−(x+1) y′ + ((2x(2x−1))/((x+1)^2 )) = 0 and x(x^2 +x+1) y′ +((x/2) +1) y = (x^2 +x+1)^(3/2)


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Question Number 173507 by JordanRoddy last updated on 12/Jul/22

$s o l v e y - (x + 1) y^{'} + \frac{2 x (2 x - 1)}{{(x + 1)}^{2}} = 0$ $and x (x^{2} + x + 1) y^{'} + (\frac{x}{2} + 1) y = {(x^{2} + x + 1)}^{\frac{3}{2}}$
Commented by JordanRoddy last updated on 12/Jul/22

$s o l v e y - (x + 1) y^{'} + \frac{2 x (2 x - 1)}{{(x + 1)}^{2}} = 0$ $and x (x^{2} + x + 1) y^{'} + (\frac{x}{2} + 1) y = {(x^{2} + x + 1)}^{\frac{3}{2}}$
Commented by mokys last updated on 12/Jul/22

$1) \frac{d y}{d x} - \frac{1}{x + 1} y = \frac{2 x (2 x - 1)}{{(x + 1)}^{3}}$ $P (x) = - \frac{1}{x + 1}, Q (x) = \frac{2 x (2 x - 1)}{{(x + 1)}^{3}}$ $I . f = e^{\int P (x) d x} = e^{\int - \frac{1}{x + 1} d x} = e^{l n \frac{1}{x + 1}} = \frac{1}{x + 1}$ $(I . f) y = \int (I . f) Q (x) d x$ $\frac{y}{x + 1} = 2 \int \frac{x (2 x - 1)}{{(x + 1)}^{4}} d x$ $l e t : m = x + 1 \to x = m - 1 \to d x = d m$ $\frac{y}{x + 1} = 2 \int \frac{(m - 1) (2 m - 3)}{m^{4}} d m$ $\frac{y}{x + 1} = 2 \int \frac{2 m^{2} - 5 m + 3}{m^{4}} d m$ $\frac{y}{x + 1} = 2 \int (\frac{2}{m^{2}} - \frac{5}{m^{3}} + \frac{3}{m^{4}}) d m$ $\frac{y}{x + 1} = - \frac{4}{m} + \frac{5}{m^{2}} - \frac{2}{m^{3}} + K$ $N o w : m = x + 1$ $∴ y = \frac{5}{x + 1} - \frac{2}{{(x + 1)}^{2}} - 4 + K (x + 1)$ $★ M o h a m m a d A l d o l a i m y ★$
Commented by Tawa11 last updated on 13/Jul/22

$Great sir .$
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