Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

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

$${Thanks}\:{so}\:{much} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com