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Question-227248

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Question Number 227248 by Spillover last updated on 10/Jan/26
Answered by breniam last updated on 10/Jan/26
z(x)=((2x+y(x)−1)/(x−2))=2+((y(x)+3)/(x−2))⇒ (z(x)−2)(x−2)−3=y(x) y′(x)=z′(x)(x−2)+z(x)−2 z′(x)(x−2)+z(x)−2=z^2 (x) z′(x)(x−2)=z^2 (x)−z(x)+2 ((z′(x))/(z^2 (x)−z(x)+2))=(1/(x−2)) =1−8<0 ∫((z′(x))/(z^2 (x)−z(x)+2))dx={t=z(x)}=∫(1/(t^2 −t+2))dt= ∫(1/(t^2 −t+(1/4)+(3/4)))dt=∫(1/((t−(1/2))^2 +(3/4)))dt= (4/3)∫(1/((((2t−1)/( (√3))))^2 +1))dt={u=((2t−1)/( (√3)))}= (2/( (√3)))∫(1/(u^2 +1))=arctan(((2z(x)−1)/( (√3))))=∫(1/(x−2))dx=ln∣x−2∣+A ((2z(x)−1)/( (√3)))=tan(ln∣x−2∣+A) z(x)=((√3)/2)tan(ln∣x−2∣+A)+(1/2) y(x)=(((√3)/2)tan(ln∣x−2∣+A)−(3/2))(x−2)−3
$${z}\left({x}\right)=\frac{\mathrm{2}{x}+{y}\left({x}\right)−\mathrm{1}}{{x}−\mathrm{2}}=\mathrm{2}+\frac{{y}\left({x}\right)+\mathrm{3}}{{x}−\mathrm{2}}\Rightarrow \\ $$$$\left({z}\left({x}\right)−\mathrm{2}\right)\left({x}−\mathrm{2}\right)−\mathrm{3}={y}\left({x}\right) \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)\left({x}−\mathrm{2}\right)+{z}\left({x}\right)−\mathrm{2} \\ $$$${z}'\left({x}\right)\left({x}−\mathrm{2}\right)+{z}\left({x}\right)−\mathrm{2}={z}^{\mathrm{2}} \left({x}\right) \\ $$$${z}'\left({x}\right)\left({x}−\mathrm{2}\right)={z}^{\mathrm{2}} \left({x}\right)−{z}\left({x}\right)+\mathrm{2} \\ $$$$\frac{{z}'\left({x}\right)}{{z}^{\mathrm{2}} \left({x}\right)−{z}\left({x}\right)+\mathrm{2}}=\frac{\mathrm{1}}{{x}−\mathrm{2}} \\ $$$$\:=\mathrm{1}−\mathrm{8}<\mathrm{0} \\ $$$$\int\frac{{z}'\left({x}\right)}{{z}^{\mathrm{2}} \left({x}\right)−{z}\left({x}\right)+\mathrm{2}}\mathrm{d}{x}=\left\{{t}={z}\left({x}\right)\right\}=\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{2}}\mathrm{d}{t}= \\ $$$$\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}\mathrm{d}{t}=\int\frac{\mathrm{1}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\mathrm{d}{t}= \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{t}=\left\{{u}=\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right\}= \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\left(\frac{\mathrm{2}{z}\left({x}\right)−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\int\frac{\mathrm{1}}{{x}−\mathrm{2}}\mathrm{d}{x}=\mathrm{ln}\mid{x}−\mathrm{2}\mid+{A} \\ $$$$\frac{\mathrm{2}{z}\left({x}\right)−\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\mathrm{tan}\left(\mathrm{ln}\mid{x}−\mathrm{2}\mid+{A}\right) \\ $$$${z}\left({x}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}\left(\mathrm{ln}\mid{x}−\mathrm{2}\mid+{A}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}\left({x}\right)=\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}\left(\mathrm{ln}\mid{x}−\mathrm{2}\mid+{A}\right)−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}−\mathrm{2}\right)−\mathrm{3} \\ $$$$ \\ $$

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