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Question Number 6086 by sanusihammed last updated on 12/Jun/16
Find the solution of the differential equation (y − x + 1)dy − (y + x + 2)dx = 0
$${Find}\:{the}\:{solution}\:{of}\:{the}\:{differential}\:{equation}\: \\ $$$$\left({y}\:−\:{x}\:+\:\mathrm{1}\right){dy}\:−\:\left({y}\:+\:{x}\:+\:\mathrm{2}\right){dx}\:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by Yozzii last updated on 12/Jun/16
M(x,y)dx+N(x,y)dy=0 M(x,y)=−y−x−2 , N(x,y)=y−x+1 ((∂M(x,y))/∂y)=−1 , ((∂N(x,y))/∂x)=−1 ∴ ((∂M(x,y))/∂y)=((∂N(x,y))/∂x) ⇒ the given differential equation is exact. ⇒∃g(x,y)∈R such that ((∂g(x,y))/∂x)=M(x,y), ((∂g(x,y))/∂y)=N(x,y) and d(g(x,y))=M(x,y)dx+N(x,y)dy. (∂g/∂x)=−y−x−2⇒g=∫(−y−x−2)dx g(x,y)=−xy−(x^2 /2)−2x+h(y).....(i) Also, (∂g/∂y)=y−x+1⇒g=∫(y−x+1)dy g(x,y)=(y^2 /2)−xy+y+f(x).....(ii) Comparing (i) and (ii) suggests that g(x,y)=−xy+(y^2 /2)+y−(x^2 /2)−2x+D with the possibility of an added constant D. Now since M(x,y)dx+N(x,y)dy=0 ⇒d(g(x,y))=0⇒g(x,y)=∫0dx=C ⇒y−2x−xy+(1/2)(x^2 +y^2 )+D=C or x^2 +y^2 +2y−4x−2xy+E=0. This is an implicit solution.
$${M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}=\mathrm{0} \\ $$$${M}\left({x},{y}\right)=−{y}−{x}−\mathrm{2}\:,\:{N}\left({x},{y}\right)={y}−{x}+\mathrm{1} \\ $$$$\frac{\partial{M}\left({x},{y}\right)}{\partial{y}}=−\mathrm{1}\:,\:\frac{\partial{N}\left({x},{y}\right)}{\partial{x}}=−\mathrm{1} \\ $$$$\therefore\:\frac{\partial{M}\left({x},{y}\right)}{\partial{y}}=\frac{\partial{N}\left({x},{y}\right)}{\partial{x}}\:\Rightarrow\:{the}\:{given} \\ $$$${differential}\:{equation}\:{is}\:{exact}. \\ $$$$\Rightarrow\exists{g}\left({x},{y}\right)\in\mathbb{R}\:{such}\:{that}\:\frac{\partial{g}\left({x},{y}\right)}{\partial{x}}={M}\left({x},{y}\right), \\ $$$$\frac{\partial{g}\left({x},{y}\right)}{\partial{y}}={N}\left({x},{y}\right)\:{and}\:{d}\left({g}\left({x},{y}\right)\right)={M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}. \\ $$$$\frac{\partial{g}}{\partial{x}}=−{y}−{x}−\mathrm{2}\Rightarrow{g}=\int\left(−{y}−{x}−\mathrm{2}\right){dx} \\ $$$${g}\left({x},{y}\right)=−{xy}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}+{h}\left({y}\right)…..\left({i}\right) \\ $$$$ \\ $$$${Also},\:\frac{\partial{g}}{\partial{y}}={y}−{x}+\mathrm{1}\Rightarrow{g}=\int\left({y}−{x}+\mathrm{1}\right){dy} \\ $$$${g}\left({x},{y}\right)=\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−{xy}+{y}+{f}\left({x}\right)…..\left({ii}\right) \\ $$$${Comparing}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{suggests}\:{that} \\ $$$${g}\left({x},{y}\right)=−{xy}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{y}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}+{D} \\ $$$${with}\:{the}\:{possibility}\:{of}\:{an}\:{added}\:{constant}\:{D}. \\ $$$${Now}\:{since}\:{M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}=\mathrm{0} \\ $$$$\Rightarrow{d}\left({g}\left({x},{y}\right)\right)=\mathrm{0}\Rightarrow{g}\left({x},{y}\right)=\int\mathrm{0}{dx}={C} \\ $$$$\Rightarrow{y}−\mathrm{2}{x}−{xy}+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+{D}={C} \\ $$$${or}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{4}{x}−\mathrm{2}{xy}+{E}=\mathrm{0}. \\ $$$${This}\:{is}\:{an}\:{implicit}\:{solution}. \\ $$$$ \\ $$
Commented by sanusihammed last updated on 13/Jun/16
Thanks so much
$${Thanks}\:{so}\:{much} \\ $$

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