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

Respuestas

Respuesta dada por: CarlosMath
0
(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\\ \\ \\
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