^(@bemath@) (1) 3^x + 3^((√x) ) = 90. find x ? (2) x (dy/dx)−(1+x)y = xy^2


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Question Number 106774 by bemath last updated on 07/Aug/20

$\overset{@ bemath @}{}$ $(1) 3^{x} + 3^{\sqrt{x}} = 90 . find x ?$ $(2) x \frac{dy}{dx} - (1 + x) y = {xy}^{2}$
	Answered by john santu last updated on 07/Aug/20

	$(1) by observe x = 4 \Rightarrow 3^{4} + 3^{\sqrt{4}} =$ $81 + 9 = 90 .^{@ JS @}$
	Answered by bobhans last updated on 07/Aug/20

	$^{≻ bobhans ≺}$ $\frac{dy}{dx} - (\frac{1 + x}{x}) y = y^{2} (Bernoulli diff eq)$ $set u = y^{- 1} \Rightarrow \frac{du}{dx} = - y^{- 2} \frac{dy}{dx}; \frac{dy}{dx} = - y^{2} \frac{du}{dx}$ $\Leftrightarrow - y^{2} \frac{du}{dx} - (\frac{1 + x}{x}) y = y^{2}$ $\frac{du}{dx} + (\frac{1 + x}{x}) u = - 1; integrating factor$ $v (x) = e^{\int (1 + \frac{1}{x}) dx} = e^{x + \ln x} = {xe}^{x}$ $u (x) = \frac{\int - {xe}^{x} dx + C}{{xe}^{x}} = \frac{- {xe}^{x} + e^{x} + C}{{xe}^{x}}$ $\frac{1}{y} = \frac{- {xe}^{x} + e^{x} + C}{{xe}^{x}} \Rightarrow y = \frac{{xe}^{x}}{C - {xe}^{x} + e^{x}}$
	Answered by abdomathmax last updated on 07/Aug/20
	$3) xy^′ −(1+x)y =xy^2 ⇒(y^′ /y^2 )−(1+(1/x))(1/y)=1 let z =(1/y) ⇒z^′ =−(y^′ /y^2 ) e⇒−z^′ −(1+(1/x))z =1 ⇒z^′ +(1+(1/x))z =−1 he→z^′ =−(1+(1/x))z ⇒(z^′ /z) =−(1+(1/x)) ⇒ ln∣z∣ =−x−ln∣x∣ +c ⇒z = k e^(−x) ×(1/(∣x∣)) solution on{x /x>0} ⇒z =k (e^(−x) /x) mvc mrthod→z^(′ ) =k^′ (e^(−x) /x) +k ×((−xe^(−x) −e^(−x) )/x^2 ) =k^′ (e^(−x) /x) −k (e^(−x) /x) −k (e^(−x) /x^2 ) e ⇒ k^′ (e^(−x) /x) −k (e^(−x) /x)−k (e^(−x) /x^2 ) +(1+(1/x))k (e^(−x) /x) =−1 ⇒k^′ =−x e^x ⇒k =−∫x e^x dx +c =−{ xe^x −∫ e^x dx} +c=−(x−1)e^x +c ⇒ z =(e^(−x) /x)( −(x−1)e^x +c) =−((x−1)/x) +((ce^(−x) )/x) we hsvr y =(1/z) ⇒ y =(x/(1−x +ce^(−x) ))$
	$3) {xy}^{^{'}} - (1 + x) y = {xy}^{2} \Rightarrow \frac{y^{^{'}}}{y^{2}} - (1 + \frac{1}{x}) \frac{1}{y} = 1$ $let z = \frac{1}{y} \Rightarrow z^{^{'}} = - \frac{y^{^{'}}}{y^{2}}$ $e \Rightarrow - z^{^{'}} - (1 + \frac{1}{x}) z = 1 \Rightarrow z^{^{'}} + (1 + \frac{1}{x}) z = - 1$ $he \to z^{^{'}} = - (1 + \frac{1}{x}) z \Rightarrow \frac{z^{^{'}}}{z} = - (1 + \frac{1}{x}) \Rightarrow$ $\ln ∣ z ∣ = - x - \ln ∣ x ∣ + c \Rightarrow z = k e^{- x} \times \frac{1}{∣ x ∣}$ $solution on {x / x > 0} \Rightarrow z = k \frac{e^{- x}}{x}$ $mvc mrthod \to z^{^{'}} = k^{^{'}} \frac{e^{- x}}{x} + k \times \frac{- {xe}^{- x} - e^{- x}}{x^{2}}$ $= k^{^{'}} \frac{e^{- x}}{x} - k \frac{e^{- x}}{x} - k \frac{e^{- x}}{x^{2}}$ $e \Rightarrow k^{^{'}} \frac{e^{- x}}{x} - k \frac{e^{- x}}{x} - k \frac{e^{- x}}{x^{2}} + (1 + \frac{1}{x}) k \frac{e^{- x}}{x} = - 1$ $\Rightarrow k^{^{'}} = - x e^{x} \Rightarrow k = - \int x e^{x} dx + c$ $= - {{xe}^{x} - \int e^{x} dx} + c = - (x - 1) e^{x} + c \Rightarrow$ $z = \frac{e^{- x}}{x} (- (x - 1) e^{x} + c) = - \frac{x - 1}{x} + \frac{{ce}^{- x}}{x}$ $we hsvr y = \frac{1}{z} \Rightarrow y = \frac{x}{1 - x + {ce}^{- x}}$
	Answered by Dwaipayan Shikari last updated on 07/Aug/20
	$(dy/dx)−(((1+x))/x)y=y^2 (1/y) (dy/dx)−((1+x)/x)=y {(1/y)=v , (dy/dx)v+y(dv/dx)=0 ,(1/y).(dy/dx)=−(1/v).(dv/dx) −(1/v).(dv/dx)−((1+x)/x)=(1/v) (dv/dx)+((1+x)/x)v=−1 I.F=e^(∫((1+x)/x)) =e^((logx+x)) =xe^x v.xe^x =−∫xe^x v.xe^x =−xe^x +e^x +C v=−1+(1/x)+(C/x)e^(−x) (1/y)=−1+(1/x)+(C/x)e^(−x) (1/y)=((−xe^x +e^x +C)/(xe^x )) y=((xe^x )/((e^x −xe^x +C)))$
	$\frac{d y}{d x} - \frac{(1 + x)}{x} y = y^{2}$ $\frac{1}{y} \frac{d y}{d x} - \frac{1 + x}{x} = y {\frac{1}{y} = v, \frac{d y}{d x} v + y \frac{d v}{d x} = 0, \frac{1}{y} . \frac{d y}{d x} = - \frac{1}{v} . \frac{d v}{d x}$ $- \frac{1}{v} . \frac{d v}{d x} - \frac{1 + x}{x} = \frac{1}{v}$ $\frac{d v}{d x} + \frac{1 + x}{x} v = - 1$ $I . F = e^{\int \frac{1 + x}{x}} = e^{(l o g x + x)} = {x e}^{x}$ $v . {x e}^{x} = - \int {x e}^{x}$ $v . {x e}^{x} = - {x e}^{x} + e^{x} + C$ $v = - 1 + \frac{1}{x} + \frac{C}{x} e^{- x}$ $\frac{1}{y} = - 1 + \frac{1}{x} + \frac{C}{x} e^{- x}$ $\frac{1}{y} = \frac{- {x e}^{x} + e^{x} + C}{{x e}^{x}}$ $y = \frac{{x e}^{x}}{(e^{x} - {x e}^{x} + C)}$
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