Question Number 90770 by ajfour last updated on 26/Apr/20
$$\left({ax}+{by}+{c}\right){dx}+\left({px}+{qy}+{r}\right){dy}=\mathrm{0} \\ $$
Answered by mr W last updated on 26/Apr/20
![(dy/dx)=−((ax+by+c)/(px+qy+r)) let x=u+u_0 , y=v+v_0 ax+by+c=a(u+u_0 )+b(v+v_0 )+c=au+bv+au_0 +bv_0 +c px+qy+r=p(u+u_0 )+q(v+v_0 )+r=pu+qv+pu_0 +qv_0 +r set au_0 +bv_0 +c=0 set pu_0 +qv_0 +r=0 ⇒u_0 =((−cq+rb)/(aq−bp)) ⇒v_0 =((−cp+ra)/(aq−bp)) (dy/dx)=(dv/du)=((au+bv)/(pu+qv)) let v=ut (dv/du)=t+u(dt/du) t+u(dt/du)=((a+bt)/(p+qt)) u(dt/du)=((a+bt)/(p+qt))−t=−((qt^2 +(p−b)t−a)/(qt+p)) ((qt+p)/(qt^2 +(p−b)t−a))dt=−(du/u) ∫((t+(p/q))/(t^2 +((p/q)−(b/q))t−(a/q)))dt=−∫(du/u) ∫((t+C)/(t^2 +At+B))dt=−∫(du/u) (1/2)ln (t^2 +At+B)+(C−(A/2))∫(dt/(t^2 +At+B))=−∫(du/u) (1/2)ln (t^2 +At+B)+((2C−A)/( (√(4B−A^2 )))) tan^(−1) ((2t+A)/( (√(4B−A^2 ))))=−ln u+K (here only case 4B−A^2 >0, for cases 4B−A^2 ≤0 similarly) ⇒(1/2)ln [(((y−v_0 )/(x−u_0 )))^2 +A(((y−v_0 )/(x−u_0 )))+B]+((2C−A)/( (√(4B−A^2 )))) tan^(−1) ((2(((y−v_0 )/(x−u_0 )))+A)/( (√(4B−A^2 ))))+ln (x−u_0 )=K](https://www.tinkutara.com/question/Q90776.png)
$$\frac{{dy}}{{dx}}=−\frac{{ax}+{by}+{c}}{{px}+{qy}+{r}} \\ $$$${let}\:{x}={u}+{u}_{\mathrm{0}} ,\:{y}={v}+{v}_{\mathrm{0}} \\ $$$${ax}+{by}+{c}={a}\left({u}+{u}_{\mathrm{0}} \right)+{b}\left({v}+{v}_{\mathrm{0}} \right)+{c}={au}+{bv}+{au}_{\mathrm{0}} +{bv}_{\mathrm{0}} +{c} \\ $$$${px}+{qy}+{r}={p}\left({u}+{u}_{\mathrm{0}} \right)+{q}\left({v}+{v}_{\mathrm{0}} \right)+{r}={pu}+{qv}+{pu}_{\mathrm{0}} +{qv}_{\mathrm{0}} +{r} \\ $$$${set}\:{au}_{\mathrm{0}} +{bv}_{\mathrm{0}} +{c}=\mathrm{0} \\ $$$${set}\:{pu}_{\mathrm{0}} +{qv}_{\mathrm{0}} +{r}=\mathrm{0} \\ $$$$\Rightarrow{u}_{\mathrm{0}} =\frac{−{cq}+{rb}}{{aq}−{bp}} \\ $$$$\Rightarrow{v}_{\mathrm{0}} =\frac{−{cp}+{ra}}{{aq}−{bp}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dv}}{{du}}=\frac{{au}+{bv}}{{pu}+{qv}} \\ $$$${let}\:{v}={ut} \\ $$$$\frac{{dv}}{{du}}={t}+{u}\frac{{dt}}{{du}} \\ $$$${t}+{u}\frac{{dt}}{{du}}=\frac{{a}+{bt}}{{p}+{qt}} \\ $$$${u}\frac{{dt}}{{du}}=\frac{{a}+{bt}}{{p}+{qt}}−{t}=−\frac{{qt}^{\mathrm{2}} +\left({p}−{b}\right){t}−{a}}{{qt}+{p}} \\ $$$$\frac{{qt}+{p}}{{qt}^{\mathrm{2}} +\left({p}−{b}\right){t}−{a}}{dt}=−\frac{{du}}{{u}} \\ $$$$\int\frac{{t}+\frac{{p}}{{q}}}{{t}^{\mathrm{2}} +\left(\frac{{p}}{{q}}−\frac{{b}}{{q}}\right){t}−\frac{{a}}{{q}}}{dt}=−\int\frac{{du}}{{u}} \\ $$$$\int\frac{{t}+{C}}{{t}^{\mathrm{2}} +{At}+{B}}{dt}=−\int\frac{{du}}{{u}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +{At}+{B}\right)+\left({C}−\frac{{A}}{\mathrm{2}}\right)\int\frac{{dt}}{{t}^{\mathrm{2}} +{At}+{B}}=−\int\frac{{du}}{{u}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +{At}+{B}\right)+\frac{\mathrm{2}{C}−{A}}{\:\sqrt{\mathrm{4}{B}−{A}^{\mathrm{2}} }}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{t}+{A}}{\:\sqrt{\mathrm{4}{B}−{A}^{\mathrm{2}} }}=−\mathrm{ln}\:{u}+{K} \\ $$$$\left({here}\:{only}\:{case}\:\mathrm{4}{B}−{A}^{\mathrm{2}} >\mathrm{0},\:{for}\:{cases}\right. \\ $$$$\left.\mathrm{4}{B}−{A}^{\mathrm{2}} \leqslant\mathrm{0}\:{similarly}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left[\left(\frac{{y}−{v}_{\mathrm{0}} }{{x}−{u}_{\mathrm{0}} }\right)^{\mathrm{2}} +{A}\left(\frac{{y}−{v}_{\mathrm{0}} }{{x}−{u}_{\mathrm{0}} }\right)+{B}\right]+\frac{\mathrm{2}{C}−{A}}{\:\sqrt{\mathrm{4}{B}−{A}^{\mathrm{2}} }}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\left(\frac{{y}−{v}_{\mathrm{0}} }{{x}−{u}_{\mathrm{0}} }\right)+{A}}{\:\sqrt{\mathrm{4}{B}−{A}^{\mathrm{2}} }}+\mathrm{ln}\:\left({x}−{u}_{\mathrm{0}} \right)={K} \\ $$
Commented by ajfour last updated on 26/Apr/20
$${Thanks}\:{Sir},\:{your}\:{presentation} \\ $$$${of}\:{every}\:{solution}\:{is}\:{beyond}\:{all} \\ $$$${praise}! \\ $$