Question Number 170795 by ali009 last updated on 31/May/22
$${solve} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{x}\left({x}−\mathrm{2}{y}\right){dy}=\mathrm{0} \\ $$
Answered by LEKOUMA last updated on 31/May/22
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}+{x}\left({x}−\mathrm{2}{y}\right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}+{x}\left({x}−\mathrm{2}{y}\right){y}^{'} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{xyy}^{'} =\mathrm{0} \\ $$$${Let}\:{y}={xt} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} {ty}^{'} =\mathrm{0} \\ $$$$\mathrm{2}+{t}^{\mathrm{2}} −\mathrm{2}{ty}^{'} =\mathrm{0} \\ $$$$−\mathrm{2}{ty}^{'} =−{t}^{\mathrm{2}} −\mathrm{2} \\ $$$${y}^{'} =\frac{−{t}^{\mathrm{2}} −\mathrm{2}}{−\mathrm{2}{t}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{t}}{\mathrm{2}}+\frac{\mathrm{1}}{{t}}=\frac{{t}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{t}} \\ $$$$\frac{{dx}}{{dy}}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{2}} \\ $$$${dx}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$$\int{dx}=\int\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$${x}=\mathrm{ln}\:\mid{t}^{\mathrm{2}} +\mathrm{2}\mid+{c} \\ $$
Commented by mr W last updated on 31/May/22
$${where}\:{is}\:{the}\:\mathrm{1}\:{gone}? \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}+{x}\left({x}−\mathrm{2}{y}\right){y}^{'} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{xyy}^{'} =\mathrm{0} \\ $$
Answered by ali009 last updated on 01/Jun/22
![it can be solved using this rule M(x,y)dx+N(x,y)dy=0 if M_y =N_x then the D.E solution is ∫M(x,y)dx + ∫N(x,y)dy − ∫(∂/∂y)(∫M(x,y)dx)dy=c the rule reference is erwin kreyszig advanved engineering mathmatics part (A) / cahpter(1) / 1.4 so [(x^2 +y^2 +1)dx+x(x−2y)dy=0]÷x^2 (1+(y^2 /x^2 )+(1/x^2 ))dx+(1−((2y)/x))dy=0 M_y =((2y)/x^2 ) N_x =((2y)/x^2 ) ∗∫(1+(y^2 /x^2 )+(1/x^2 ))dx=x−(y^2 /x)−(1/x) ∗∫(1−((2y)/x))dy=y−(y^2 /x) ∗∫−((2y)/x)dy=−(y^2 /x) x+y−((y^2 +1)/x)=c](https://www.tinkutara.com/question/Q170828.png)
$${it}\:{can}\:{be}\:{solved}\:{using}\:{this}\:{rule} \\ $$$${M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}=\mathrm{0} \\ $$$${if}\:{M}_{{y}} ={N}_{{x}} \\ $$$${then}\:{the}\:{D}.{E}\:{solution}\:{is}\: \\ $$$$\int{M}\left({x},{y}\right){dx}\:+\:\int{N}\left({x},{y}\right){dy}\:−\:\int\frac{\partial}{\partial{y}}\left(\int{M}\left({x},{y}\right){dx}\right){dy}={c} \\ $$$${the}\:{rule}\:{reference}\:{is} \\ $$$$\:{erwin}\:{kreyszig}\:{advanved}\:{engineering}\:{mathmatics} \\ $$$${part}\:\left({A}\right)\:/\:{cahpter}\left(\mathrm{1}\right)\:/\:\mathrm{1}.\mathrm{4}\: \\ $$$$ \\ $$$${so} \\ $$$$\left[\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{x}\left({x}−\mathrm{2}{y}\right){dy}=\mathrm{0}\right]\boldsymbol{\div}{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}+\left(\mathrm{1}−\frac{\mathrm{2}{y}}{{x}}\right){dy}=\mathrm{0} \\ $$$${M}_{{y}} =\frac{\mathrm{2}{y}}{{x}^{\mathrm{2}} }\:\:\:\:\:\:\:{N}_{{x}} =\frac{\mathrm{2}{y}}{{x}^{\mathrm{2}} } \\ $$$$\ast\int\left(\mathrm{1}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}={x}−\frac{{y}^{\mathrm{2}} }{{x}}−\frac{\mathrm{1}}{{x}} \\ $$$$\ast\int\left(\mathrm{1}−\frac{\mathrm{2}{y}}{{x}}\right){dy}={y}−\frac{{y}^{\mathrm{2}} }{{x}} \\ $$$$\ast\int−\frac{\mathrm{2}{y}}{{x}}{dy}=−\frac{{y}^{\mathrm{2}} }{{x}} \\ $$$${x}+{y}−\frac{{y}^{\mathrm{2}} +\mathrm{1}}{{x}}={c} \\ $$$$ \\ $$$$ \\ $$