Solve-the-differential-equation-xy-y-2x-1-x-1-2-0-

Question Number 99260 by Ar Brandon last updated on 19/Jun/20

Solve the differential equation

{xy}^{'} - y + \frac{2 x + 1}{{(x + 1)}^{2}} = 0

Answered by Mr.D.N. last updated on 19/Jun/20

Commented by Ar Brandon last updated on 20/Jun/20

Thank you ��

Answered by mathmax by abdo last updated on 19/Jun/20

{xy}^{^{'}} - y + \frac{2 x + 1}{{(x + 1)}^{2}} = 0 (he) \to {xy}^{^{'}} - y = 0 \Rightarrow \frac{y^{^{'}}}{y} = \frac{1}{x} \Rightarrow \ln ∣ y ∣ = \ln ∣ x ∣ + c \Rightarrow

y (x) = k ∣ x ∣ let determine slution on] 0, + \infty [\Rightarrow y (x) = kx

mvc method \to y^{^{'}} = k^{^{'}} x + k

(e) \Rightarrow k^{^{'}} x^{2} + kx - kx = - \frac{2 x + 1}{{(x + 1)}^{2}} \Rightarrow k^{^{'}} = - \frac{2 x + 1}{x^{2} {(x + 1)}^{2}} \Rightarrow k (x) = - \int \frac{2 x + 1}{x^{2} {(x + 1)}^{2}} dx + c

let decompose F (x) = \frac{2 x + 1}{x^{2} {(x + 1)}^{2}} we have \frac{1}{x^{2}} - \frac{1}{{(x + 1)}^{2}} = \frac{{(x + 1)}^{2} - x^{2}}{x^{2} {(x + 1)}^{2}}

= \frac{2 x + 1}{x^{2} {(x + 1)}^{2}} = F (x) \Rightarrow k (x) = - \int \frac{dx}{x^{2}} + \int \frac{dx}{{(x + 1)}^{2}} = \frac{1}{x} - \frac{1}{x + 1} + c \Rightarrow

y (x) = xk (x) = x (\frac{1}{x} - \frac{1}{x + 1} + c) = 1 - \frac{x}{x + 1} + c = \frac{1}{x + 1} + c

Commented by mathmax by abdo last updated on 19/Jun/20

sorry y (x) = \frac{1}{x + 1} + cx

Commented by Ar Brandon last updated on 20/Jun/20

Thanks

Commented by mathmax by abdo last updated on 20/Jun/20

you are welcome