Question

ayudenme a resolver este ejercio de ecuaciones exactas 
(bx^2y+3xy^2)dx+x^2(x+3y)dy=0

Answer 1

(1) Veamos si es exacta

$P(x,y)=bx^2y+3xy^2~~\&~~Q(x,y)=x^2(x+3y)\\ \\ P_y=bx^2+6xy~~\&~~Q_x=3x^2+6xy$

No es exacta

(2) resolvámosla de otra manera

$(bx^2y+3xy^2)dx+x^2(x+3y)dy=0\\ \\ y(bx+3y)dx+x(x+3y)dy=0\\ \\ \\ *\texttt{Sea }x=yz\to dx=y~dz+z~dy\\ \\ y(byz+3y)(y~dz+z~dy)+yz(yz+3y)dy=0\\ \\ \\ (bz+3)(y~dz+z~dy)+z(z+3)dy=0\\ \\ \\ (bz+3)y~dz+z((b+1)z+6)dy=0\\ \\ \dfrac{bz+3}{z((b+1)z+6)}+\dfrac{dy}{y}=0\\ \\ \\ \displaystyle \int\dfrac{bz+3}{z((b+1)z+6)}dz+\int\dfrac{dy}{y}=0\\ \\ \\ \int\dfrac{bz+3}{z((b+1)z+6)}dz+\ln|y|=0$


$\displaystyle \dfrac{b-1}{2(b+1)}\int \dfrac{dz}{z+\dfrac{6}{b+1}}+\dfrac{1}{2}\int\dfrac{dz}{z}+\ln|y|=C\\ \\ \\ \dfrac{b-1}{2(b+1)}\ln\left|z+\dfrac{6}{b+1}\right|+\dfrac{1}{2}\ln|z|+\ln|y|=C\\ \\ \\ \left(z+\dfrac{6}{b+1}\right)^{\dfrac{b-1}{2(b+1)}}\cdot\sqrt{z}\cdot y=C_1\ \textgreater \ 0\\ \\ \\ \left(z+\dfrac{6}{b+1}\right)^{\dfrac{b-1}{b+1}}\cdot|z|\cdot y^2=C_1^2$


$\left(\dfrac{x}{y}+\dfrac{6}{b+1}\right)^{\dfrac{b-1}{b+1}}\cdot\left|\dfrac{x}{y}\right|\cdot y^2=C_1^2\\ \\ \\ \left(\dfrac{x}{y}+\dfrac{6}{b+1}\right)^{\dfrac{b-1}{b+1}}|xy|=C_1^2$