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Question Number 196934 by Amidip last updated on 03/Sep/23

Answered by aleks041103 last updated on 04/Sep/23

$$M(x,y)dx+N(x,y)dy=0$$$$if{\partial }_{y}M={\partial }_{x}N,then\mathrm{\exists }F(x,y):$$$$M(x,y)dx+N(x,y)dy=dF=0\Rightarrow F=const.$$$$inourcase{\partial }_{y}M\ne {\partial }_{x}N$$$$findintegratingfactorq(x,y),s.t.$$$${\partial }_y\left(qM\right)={\partial }_x\left(qN\right)$$$$1)supposeq=q\left(x\right)$$$$\Rightarrow q{\partial }_{y}M=q{\partial }_{x}N+q{}^{\prime }N\Rightarrow {Nq}^{\prime }=q({\partial }_{y}M-{\partial }_{x}N)$$$$inourcase:$$$$(4x^3{y}^3-3x^2)q^{\prime }=q(24x^2{y}^3+6x-12x^2{y}^3+6x)$$$$\Rightarrow q^{\prime }/q=\frac{12x^2{y}^3+12x}{4x^3{y}^3-3x^2}\ne f\left(x\right)$$$$2)supposeq=q\left(y\right)$$$$q{\partial }_{y}M+{Mq}^{\prime }=q{\partial }_{x}N\Rightarrow \frac{q^{\prime }}{q}=-\frac{{\partial }_{y}M-{\partial }_{x}N}{M}$$$$\frac{q^{\prime }}{q}=-\frac{12(x^2{y}^3+x)}{6(x^2{y}^4+xy)}=\frac{-2}{y}$$$$\Rightarrow d(ln\left(q\right)+2ln\left(y\right))=0\Rightarrow q=\frac{1}{y^2}$$$$\Rightarrow 6(x^2{y}^4+xy)dx+(4x^3{y}^3-3x^2)dy=0$$$$\Rightarrow (6x^2{y}^2+\frac{6x}{y})dx+(4x^{3}y-\frac{3x^2}{y^2})dy=dF=0$$$${\partial }_{x}F=M=6x^2{y}^2+\frac{6x}{y}\Rightarrow F=2x^3{y}^2+\frac{3x^2}{y}+f\left(y\right)$$$$\Rightarrow {\partial }_{y}F=4x^{3}y-\frac{3x^2}{y^2}+f{}^{\prime }=N=4x^{3}y-\frac{3x^2}{y^2}$$$$\Rightarrow f{}^{\prime }=0\Rightarrow f=c=const.$$$$\Rightarrow F(x,y)=2x^3{y}^2+\frac{3x^2}{y}=const.$$$$\Rightarrow Thesolutionsarethecurvesgivenby:$$$$2x^3{y}^3+3x^2-\alpha y=0,\mathrm{\forall }\alpha \in \mathbb{R}$$

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