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Question-196934

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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 ∂_y M=∂_x N, then ∃F(x,y): M(x,y)dx+N(x,y)dy=dF=0⇒F=const. in our case ∂_y M≠∂_x N find integrating factor q(x,y), s.t. ∂_y (qM)=∂_x (qN) 1)suppose q=q(x) ⇒q∂_y M=q∂_x N+q ′ N⇒Nq′=q(∂_y M−∂_x N) in our case: (4x^3 y^3 −3x^2 )q′=q(24x^2 y^3 +6x−12x^2 y^3 +6x) ⇒q′/q=((12x^2 y^3 +12x)/(4x^3 y^3 −3x^2 ))≠f(x) 2)suppose q=q(y) q∂_y M+Mq′=q∂_x N⇒((q′)/q)=−((∂_y M−∂_x N)/M) ((q′)/q)=−((12(x^2 y^3 +x))/(6(x^2 y^4 +xy)))=((−2)/y) ⇒d(ln(q)+2ln(y))=0⇒q=(1/y^2 ) ⇒6(x^2 y^4 +xy)dx+(4x^3 y^3 −3x^2 )dy=0 ⇒(6x^2 y^2 +((6x)/y))dx+(4x^3 y−((3x^2 )/y^2 ))dy=dF=0 ∂_x F=M=6x^2 y^2 +((6x)/y)⇒F=2x^3 y^2 +((3x^2 )/y)+f(y) ⇒∂_y F=4x^3 y−((3x^2 )/y^2 )+f ′=N=4x^3 y−((3x^2 )/y^2 ) ⇒f ′=0⇒f=c=const. ⇒F(x,y)=2x^3 y^2 +((3x^2 )/y)=const. ⇒The solutions are the curves given by: 2x^3 y^3 +3x^2 −αy=0, ∀α∈R
$${M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}=\mathrm{0} \\ $$$${if}\:\:\partial_{{y}} {M}=\partial_{{x}} {N},\:{then}\:\exists{F}\left({x},{y}\right): \\ $$$${M}\left({x},{y}\right){dx}+{N}\left({x},{y}\right){dy}={dF}=\mathrm{0}\Rightarrow{F}={const}. \\ $$$${in}\:{our}\:{case}\:\partial_{{y}} {M}\neq\partial_{{x}} {N} \\ $$$${find}\:{integrating}\:{factor}\:{q}\left({x},{y}\right),\:{s}.{t}. \\ $$$$\partial_{{y}} \left({qM}\right)=\partial_{{x}} \left({qN}\right) \\ $$$$\left.\mathrm{1}\right){suppose}\:{q}={q}\left({x}\right) \\ $$$$\Rightarrow{q}\partial_{{y}} {M}={q}\partial_{{x}} {N}+{q}\:'\:{N}\Rightarrow{Nq}'={q}\left(\partial_{{y}} {M}−\partial_{{x}} {N}\right) \\ $$$${in}\:{our}\:{case}: \\ $$$$\left(\mathrm{4}{x}^{\mathrm{3}} {y}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} \right){q}'={q}\left(\mathrm{24}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +\mathrm{6}{x}−\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +\mathrm{6}{x}\right) \\ $$$$\Rightarrow{q}'/{q}=\frac{\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +\mathrm{12}{x}}{\mathrm{4}{x}^{\mathrm{3}} {y}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} }\neq{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){suppose}\:{q}={q}\left({y}\right) \\ $$$${q}\partial_{{y}} {M}+{Mq}'={q}\partial_{{x}} {N}\Rightarrow\frac{{q}'}{{q}}=−\frac{\partial_{{y}} {M}−\partial_{{x}} {N}}{{M}} \\ $$$$\frac{{q}'}{{q}}=−\frac{\mathrm{12}\left({x}^{\mathrm{2}} {y}^{\mathrm{3}} +{x}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} {y}^{\mathrm{4}} +{xy}\right)}=\frac{−\mathrm{2}}{{y}} \\ $$$$\Rightarrow{d}\left({ln}\left({q}\right)+\mathrm{2}{ln}\left({y}\right)\right)=\mathrm{0}\Rightarrow{q}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{6}\left({x}^{\mathrm{2}} {y}^{\mathrm{4}} +{xy}\right){dx}+\left(\mathrm{4}{x}^{\mathrm{3}} {y}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{6}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\frac{\mathrm{6}{x}}{{y}}\right){dx}+\left(\mathrm{4}{x}^{\mathrm{3}} {y}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\right){dy}={dF}=\mathrm{0} \\ $$$$\partial_{{x}} {F}={M}=\mathrm{6}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\frac{\mathrm{6}{x}}{{y}}\Rightarrow{F}=\mathrm{2}{x}^{\mathrm{3}} {y}^{\mathrm{2}} +\frac{\mathrm{3}{x}^{\mathrm{2}} }{{y}}+{f}\left({y}\right) \\ $$$$\Rightarrow\partial_{{y}} {F}=\mathrm{4}{x}^{\mathrm{3}} {y}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{f}\:'={N}=\mathrm{4}{x}^{\mathrm{3}} {y}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}\:'=\mathrm{0}\Rightarrow{f}={c}={const}. \\ $$$$\Rightarrow{F}\left({x},{y}\right)=\mathrm{2}{x}^{\mathrm{3}} {y}^{\mathrm{2}} +\frac{\mathrm{3}{x}^{\mathrm{2}} }{{y}}={const}. \\ $$$$\Rightarrow{The}\:{solutions}\:{are}\:{the}\:{curves}\:{given}\:{by}: \\ $$$$\mathrm{2}{x}^{\mathrm{3}} {y}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\alpha{y}=\mathrm{0},\:\forall\alpha\in\mathbb{R} \\ $$

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