solve y′ =2y^2 +y and y(o)=1


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Question Number 57414 by Abdo msup. last updated on 03/Apr/19

$s o l v e y^{'} = 2 y^{2} + y a n d y (o) = 1$
Commented by kaivan.ahmadi last updated on 03/Apr/19
$first we notice the integral ∫(dx/(2x^2 +x))=∫(dx/(x(2x+1)))= (A/x)+(B/(2x+1))=(((2A+B)x+A)/(x(2x+1)))=(1/(x(2x+1)))⇒ { ((2A+B=0)),((A=1)) :}⇒A=1,B=−2 ∫(dx/x)−∫((2dx)/(2x+1))=lnx−ln(2x+1)=ln((x/(2x+1))) now we solve the problem (dy/dx)=2y^2 +y⇒dx=(dy/(2y^2 +y))⇒ ∫dx=∫(dy/(2y^2 +y))⇒x=ln((y/(2y+1)))⇒ (y/(2y+1))=e^x ⇒2ye^x +e^x =y⇒ y(2e^x −1)=−e^x ⇒y=(e^x /(1−2e^x ))+C on the other hand y(0)=1 so 1=−1+C⇒C=2⇒ y=(e^x /(1−2e^x ))+2$
$f i r s t w e n o t i c e t h e i n t e g r a l$ $\int \frac{d x}{2 x^{2} + x} = \int \frac{d x}{x (2 x + 1)} =$ $\frac{A}{x} + \frac{B}{2 x + 1} = \frac{(2 A + B) x + A}{x (2 x + 1)} = \frac{1}{x (2 x + 1)} \Rightarrow$ ${\begin{cases} 2 A + B = 0 \\ A = 1 \end{cases} \Rightarrow A = 1, B = - 2$ $\int \frac{d x}{x} - \int \frac{2 d x}{2 x + 1} = l n x - l n (2 x + 1) = l n (\frac{x}{2 x + 1})$ $n o w w e s o l v e t h e p r o b l e m$ $\frac{d y}{d x} = 2 y^{2} + y \Rightarrow d x = \frac{d y}{2 y^{2} + y} \Rightarrow$ $\int d x = \int \frac{d y}{2 y^{2} + y} \Rightarrow x = l n (\frac{y}{2 y + 1}) \Rightarrow$ $\frac{y}{2 y + 1} = e^{x} \Rightarrow 2 {y e}^{x} + e^{x} = y \Rightarrow$ $y (2 e^{x} - 1) = - e^{x} \Rightarrow y = \frac{e^{x}}{1 - 2 e^{x}} + C$ $o n t h e o t h e r h a n d y (0) = 1 s o$ $1 = - 1 + C \Rightarrow C = 2 \Rightarrow$ $y = \frac{e^{x}}{1 - 2 e^{x}} + 2$
Commented by maxmathsup by imad last updated on 04/Apr/19

$l e t u s e t h e c h a n g e m e n t y = \frac{1}{z} \Rightarrow y^{^{'}} = - \frac{z^{^{'}}}{z^{2}} a n d (e d) \Rightarrow - \frac{z^{^{'}}}{z^{2}} = \frac{2}{z^{2}} + \frac{1}{z} \Rightarrow$ $- z^{^{'}} = 2 + z \Rightarrow z^{^{'}} + z + 2 = 0 y (0) = 1 \Rightarrow \frac{1}{z (0)} = 1 \Rightarrow z (0) = 1$ $(h e) \Rightarrow z^{^{'}} = - z \Rightarrow - \frac{z^{^{'}}}{z} = 1 \Rightarrow - l n ∣ z ∣= x + c_{0} \Rightarrow z (x) = k e^{- x}$ $m v c m e t h o d g i v e z^{^{'}} = k^{^{'}} e^{- x} - k e^{- x} = (k^{^{'}} - k) e^{- x} a n d (e) \Rightarrow$ $(k^{^{'}} - k) e^{- x} + {k e}^{- x} + 2 = 0 \Rightarrow k^{^{'}} e^{- x} = - 2 \Rightarrow k^{^{'}} = - 2 e^{x} \Rightarrow$ $k = - 2 e^{x} + λ \Rightarrow z (x) = (- 2 e^{x} + λ) e^{- x} = - 2 + λ e^{- x} \Rightarrow$ $y (x) = \frac{1}{z (x)} = \frac{1}{λ e^{- x} - 2}$ $y (o) = 1 \Rightarrow \frac{1}{λ - 2} = 1 \Rightarrow λ - 2 = 1 \Rightarrow λ = 3 \Rightarrow y (x) = \frac{1}{3 e^{- x} - 2} .$
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