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Question Number 103553 by byaw last updated on 15/Jul/20

	Answered by mathmax by abdo last updated on 16/Jul/20
	$2y^(′′) −4y^′ −6y =0 ⇒y^(′′) −2y^′ −3y =0 →r^2 −2r +3 =0 →Δ^′ =1+3 =4 ⇒r_1 =1+2 =3 and r_2 =1−2 =−1 ⇒ y =α e^(−x) +βe^(3x) y(o)=3 ⇒α+β =3 y^′ =−α e^(−x) +3β e^(3x) so y^′ (0) =4 ⇒−α+3β =4 we get the system { ((α+β =3)),((−α+3β =4 ⇒4β =7 ⇒β =(7/4))) :} α=3−β =3−(7/4) =(5/4) ⇒y(x) =(5/4)e^(−x) +(7/4)e^(3x)$
	${2 y}^{^{″}} - {4 y}^{^{'}} - 6 y = 0 \Rightarrow y^{^{″}} - {2 y}^{^{'}} - 3 y = 0$ $\to r^{2} - 2 r + 3 = 0 \to Δ^{^{'}} = 1 + 3 = 4 \Rightarrow r_{1} = 1 + 2 = 3 and r_{2} = 1 - 2 = - 1 \Rightarrow$ $y = α e^{- x} + β e^{3 x}$ $y (o) = 3 \Rightarrow α + β = 3$ $y^{^{'}} = - α e^{- x} + 3 β e^{3 x} so y^{^{'}} (0) = 4 \Rightarrow - α + 3 β = 4 we get the system$ ${\begin{cases} α + β = 3 \\ - α + 3 β = 4 \Rightarrow 4 β = 7 \Rightarrow β = \frac{7}{4} \end{cases}$ $α = 3 - β = 3 - \frac{7}{4} = \frac{5}{4} \Rightarrow y (x) = \frac{5}{4} e^{- x} + \frac{7}{4} e^{3 x}$
	Answered by bobhans last updated on 16/Jul/20

	$(3) \frac{\partial M}{\partial y} = - \frac{x}{y^{2}} . e^{x / y}$ $\frac{\partial N}{\partial x} = \frac{1}{y} e^{x / y} (1 - \frac{x}{y}) - \frac{1}{y} . e^{x / y} = - \frac{x}{y^{2}} . e^{x / y}$ $b e c a u s e \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} s o t h i s i s e x a c t d i f f$ $e q u a t i o n$
	Commented by byaw last updated on 16/Jul/20

	$Thank you soo much .$
	Answered by Dwaipayan Shikari last updated on 16/Jul/20

	$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$ $\frac{d y}{d x} = \frac{v^{2} + 1}{2 v} [v y = x, \frac{d y}{d x} v + \frac{d v}{d x} y$ $\frac{d v}{d x} . y = \frac{v^{2} + 1}{2}$ $\frac{d v}{d x} . \frac{x}{v} = \frac{v^{2} + 1}{2}$ $\frac{2 d v}{v (v^{2} + 1)} = \frac{d x}{x}$ $- \int \frac{- \frac{2}{v^{3}}}{1 + \frac{1}{v^{2}}} d v = l o g x + C$ $- l o g (1 + \frac{1}{v^{2}}) = l o g x + C$ $2 l o g v - l o g (v^{2} + 1) = l o g x + C$ $2 l o g x - 2 l o g y - l o g (x^{2} + y^{2}) + 2 l o g y = l o g x + {l o g C}_{1}$ $l o g (\frac{x}{x^{2} + y^{2}} . \frac{1}{C_{1}}) = 0$ $x = C_{1} (x^{2} + y^{2})$
	Commented by byaw last updated on 16/Jul/20

	$Thank you . I am greatful$
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