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Question Number 168518 by LEKOUMA last updated on 12/Apr/22

$$R\acute{esoudre}{l}^{\prime }\acute{equation}au\mathrm{x}\mathrm{differentiel}les$$$$\mathrm{totales}$$$$2xydx+(x^2-y^2)dy=0$$

Commented by mokys last updated on 12/Apr/22

$$M(x,y)=2xy\to M_y=2x$$$$N(x,y)=x^2-y^2\to N_x=2x$$$$$$$$M_y=N_x\to theequationisexact$$$$$$$$f=\int M(x,y)dx=\int \left(2xy\right)dx=x^{2}y+A\left(y\right)$$$$$$$$\because f_y=N$$$$$$$$\therefore x^2+A{}^{{}^{\prime }}\left(y\right)=x^2-y^2\Rightarrow A{}^{{}^{\prime }}\left(y\right)=-y^2\Rightarrow A\left(y\right)=-\frac{y^3}{3}+C$$$$$$$$\therefore x^{2}y-\frac{y^3}{3}=C$$$$$$$$\left(aldolaimy\right)$$

Answered by alephzero last updated on 12/Apr/22

$$please,translatetoenglish$$

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