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Question Number 127354 by liberty last updated on 29/Dec/20
Answered by som(math1967) last updated on 29/Dec/20
1. let x=X+h y=Y+k (dx/dX)=1 (dy/dY)=1 (dy/dx)=(dy/dY) .(dY/dX).(dX/dx)=1.(dY/dX).1=(dY/dX) ∴(dY/dX)=((X+h+Y+k−3)/(X+h−Y−k−1)) (dY/dX)=((X+Y+h+k−3)/(X−Y+h−k−1)) now h+k−3=0 h−k−1=0 ⇒h=2 ,k=1 (dY/dX)=((X+Y)/(X−Y)) (dY/dX)=((1+(Y/X))/(1−(Y/X))) let Y=vX⇒(dY/dX)=v+(dv/dX)X v+(dv/dX)X=((1+v)/(1−v)) (dv/dX)X=((1+v−v+v^2 )/(1−v)) ∫((1−v)/(1+v^2 ))dv=∫(dX/X) tan^(−1) v−(1/2)log∣1+v^2 ∣=logX+Logc tan^(−1) ((Y/X))−(1/2)log∣1+(Y^2 /X^2 )∣=log∣Xc∣ tan^(−1) ((y−1)/(x−2))−(1/2)log∣(((x−2)^2 +(y−1)^2 )/((x−2)^2 ))∣=log∣(x−2)c∣
$$\mathrm{1}.\:{let}\:{x}={X}+{h}\:\:\:{y}={Y}+{k} \\ $$$$\frac{{dx}}{{dX}}=\mathrm{1}\:\:\:\:\frac{{dy}}{{dY}}=\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dY}}\:.\frac{{dY}}{{dX}}.\frac{{dX}}{{dx}}=\mathrm{1}.\frac{{dY}}{{dX}}.\mathrm{1}=\frac{{dY}}{{dX}} \\ $$$$\therefore\frac{{dY}}{{dX}}=\frac{{X}+{h}+{Y}+{k}−\mathrm{3}}{{X}+{h}−{Y}−{k}−\mathrm{1}} \\ $$$$\frac{{dY}}{{dX}}=\frac{{X}+{Y}+{h}+{k}−\mathrm{3}}{{X}−{Y}+{h}−{k}−\mathrm{1}} \\ $$$${now}\:{h}+{k}−\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:{h}−{k}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{h}=\mathrm{2}\:\:,{k}=\mathrm{1} \\ $$$$\frac{{dY}}{{dX}}=\frac{{X}+{Y}}{{X}−{Y}} \\ $$$$\frac{{dY}}{{dX}}=\frac{\mathrm{1}+\frac{{Y}}{{X}}}{\mathrm{1}−\frac{{Y}}{{X}}} \\ $$$${let}\:{Y}={vX}\Rightarrow\frac{{dY}}{{dX}}={v}+\frac{{dv}}{{dX}}{X} \\ $$$${v}+\frac{{dv}}{{dX}}{X}=\frac{\mathrm{1}+{v}}{\mathrm{1}−{v}} \\ $$$$\frac{{dv}}{{dX}}{X}=\frac{\mathrm{1}+{v}−{v}+{v}^{\mathrm{2}} }{\mathrm{1}−{v}} \\ $$$$\:\int\frac{\mathrm{1}−{v}}{\mathrm{1}+{v}^{\mathrm{2}} }{dv}=\int\frac{{dX}}{{X}} \\ $$$$ \\ $$$${tan}^{−\mathrm{1}} {v}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mid\mathrm{1}+{v}^{\mathrm{2}} \mid={logX}+{Logc} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{Y}}{{X}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{log}\mid\mathrm{1}+\frac{{Y}^{\mathrm{2}} }{{X}^{\mathrm{2}} }\mid={log}\mid{Xc}\mid \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{{y}−\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mid\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\mid={log}\mid\left({x}−\mathrm{2}\right){c}\mid \\ $$
Answered by bemath last updated on 29/Dec/20
(1) make the substitution → { ((x=x_1 +h)),((y=y_1 +k)) :} ⇒ (dy_1 /dx_1 ) = ((x_1 +y_1 +h+k−3)/(x_1 −y_1 +h−k−1)) solving the set of two eq { ((h+k−3=0)),((h−k−1=0)) :} we find { ((h=2)),((k=1)) :}. As result (dy_1 /dx_1 ) = ((x_1 +y_1 )/(x_1 −y_1 )) let (y_1 /x_1 ) = v ; y_1 =vx_1 and (dy_1 /dx_1 ) = v + x_1 (dv/dx_1 ) v + x_1 (dv/dx_1 ) = ((1+v)/(1−v)) ∧ ((1−v)/(1+v^2 )) dv = (dx_1 /x_1 ) ∫ ((1−v)/(1+v^2 )) dv = ln Cx_1 arctan v−(1/2) ln (1+v^2 ) = ln Cx_1 arctan v = ln Cx_1 (√(1+v^2 )) Cx_1 (√(1+v^2 )) = e^(arctan v ) C(x−2)(√(1+(((y−1)/(x−2)))^2 )) = e^(arctan (((y−1)/(x−2)))) C(√((x−2)^2 +(y−1)^2 )) = e^(arctan (((y−1)/(x−2))))
$$\left(\mathrm{1}\right)\:{make}\:{the}\:{substitution}\: \\ $$$$\:\rightarrow\begin{cases}{{x}={x}_{\mathrm{1}} +{h}}\\{{y}={y}_{\mathrm{1}} +{k}}\end{cases}\:\Rightarrow\:\frac{{dy}_{\mathrm{1}} }{{dx}_{\mathrm{1}} }\:=\:\frac{{x}_{\mathrm{1}} +{y}_{\mathrm{1}} +{h}+{k}−\mathrm{3}}{{x}_{\mathrm{1}} −{y}_{\mathrm{1}} +{h}−{k}−\mathrm{1}}\: \\ $$$${solving}\:{the}\:{set}\:{of}\:{two}\:{eq}\:\begin{cases}{{h}+{k}−\mathrm{3}=\mathrm{0}}\\{{h}−{k}−\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$${we}\:{find}\:\begin{cases}{{h}=\mathrm{2}}\\{{k}=\mathrm{1}}\end{cases}.\:{As}\:{result}\:\frac{{dy}_{\mathrm{1}} }{{dx}_{\mathrm{1}} }\:=\:\frac{{x}_{\mathrm{1}} +{y}_{\mathrm{1}} }{{x}_{\mathrm{1}} −{y}_{\mathrm{1}} } \\ $$$${let}\:\frac{{y}_{\mathrm{1}} }{{x}_{\mathrm{1}} }\:=\:{v}\:;\:{y}_{\mathrm{1}} ={vx}_{\mathrm{1}} \:{and}\:\frac{{dy}_{\mathrm{1}} }{{dx}_{\mathrm{1}} }\:=\:{v}\:+\:{x}_{\mathrm{1}} \:\frac{{dv}}{{dx}_{\mathrm{1}} } \\ $$$$\:{v}\:+\:{x}_{\mathrm{1}} \:\frac{{dv}}{{dx}_{\mathrm{1}} }\:=\:\frac{\mathrm{1}+{v}}{\mathrm{1}−{v}}\:\wedge\:\frac{\mathrm{1}−{v}}{\mathrm{1}+{v}^{\mathrm{2}} }\:{dv}\:=\:\frac{{dx}_{\mathrm{1}} }{{x}_{\mathrm{1}} } \\ $$$$\int\:\frac{\mathrm{1}−{v}}{\mathrm{1}+{v}^{\mathrm{2}} }\:{dv}\:=\:\mathrm{ln}\:{Cx}_{\mathrm{1}} \\ $$$$\:\mathrm{arctan}\:\:{v}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\left(\mathrm{1}+{v}^{\mathrm{2}} \right)\:=\:\mathrm{ln}\:{Cx}_{\mathrm{1}} \\ $$$$\:\mathrm{arctan}\:{v}\:=\:\mathrm{ln}\:{Cx}_{\mathrm{1}} \sqrt{\mathrm{1}+{v}^{\mathrm{2}} } \\ $$$$\:{Cx}_{\mathrm{1}} \:\sqrt{\mathrm{1}+{v}^{\mathrm{2}} }\:=\:{e}^{\mathrm{arctan}\:{v}\:} \\ $$$$\:{C}\left({x}−\mathrm{2}\right)\sqrt{\mathrm{1}+\left(\frac{{y}−\mathrm{1}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} }\:=\:{e}^{\mathrm{arctan}\:\left(\frac{{y}−\mathrm{1}}{{x}−\mathrm{2}}\right)} \: \\ $$$${C}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:{e}^{\mathrm{arctan}\:\left(\frac{{y}−\mathrm{1}}{{x}−\mathrm{2}}\right)} \: \\ $$
Answered by bemath last updated on 29/Dec/20
(3) (dy/dx)+(−(2/(x+1)))y = (x+1)^3 putting y = uv ⇒(dy/dx) = u (dv/dx)+ v (du/dx) putting the expression (dy/dx) into the original equation u (dv/dx)+ v (du/dx) + (−(2/(x+1)))uv = (x+1)^3 u((dv/dx) − (2/(x+1))v)+v (du/dx) = (x+1)^3 to determine v we get (dv/dx)−((2v)/(x+1)) = 0 (dv/v) = ((2dx)/(x+1)) ⇒ v = (x+1)^2 the we get (x+1)^2 (du/dx) = (x+1)^3 ⇒ (du/dx) = x+1 and u = (1/2)(x+1)^2 +C thus the complete integral of the given equation y = (((x+1)^4 )/2)+ C(x+1)^2
$$\left(\mathrm{3}\right)\:\frac{{dy}}{{dx}}+\left(−\frac{\mathrm{2}}{{x}+\mathrm{1}}\right){y}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:{putting}\:{y}\:=\:{uv}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{u}\:\frac{{dv}}{{dx}}+\:{v}\:\frac{{du}}{{dx}} \\ $$$${putting}\:{the}\:{expression}\:\frac{{dy}}{{dx}}\:{into}\:{the} \\ $$$${original}\:{equation}\: \\ $$$$\:{u}\:\frac{{dv}}{{dx}}+\:{v}\:\frac{{du}}{{dx}}\:+\:\left(−\frac{\mathrm{2}}{{x}+\mathrm{1}}\right){uv}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:{u}\left(\frac{{dv}}{{dx}}\:−\:\frac{\mathrm{2}}{{x}+\mathrm{1}}{v}\right)+{v}\:\frac{{du}}{{dx}}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:{to}\:{determine}\:{v}\:{we}\:{get}\:\frac{{dv}}{{dx}}−\frac{\mathrm{2}{v}}{{x}+\mathrm{1}}\:=\:\mathrm{0} \\ $$$$\:\frac{{dv}}{{v}}\:=\:\frac{\mathrm{2}{dx}}{{x}+\mathrm{1}}\:\Rightarrow\:{v}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${the}\:{we}\:{get}\: \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:\frac{{du}}{{dx}}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} \:\Rightarrow\:\frac{{du}}{{dx}}\:=\:{x}+\mathrm{1} \\ $$$$\:{and}\:{u}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{C}\: \\ $$$${thus}\:{the}\:{complete}\:{integral}\:{of}\:{the} \\ $$$${given}\:{equation}\:{y}\:=\:\frac{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{2}}+\:{C}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \: \\ $$$$ \\ $$
Answered by liberty last updated on 29/Dec/20

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